Question 26: 2.7^n+3.5^n-5 is divisible by 24 for all n \in N

Answer:

Let P(n) : 2.7^n+3.5^n-5 is divisible by 24 for all n \in N

For n = 1 , L.H.S = 2 \times 7 + 3 \times 5 - 5 = 24  which is divisible by 24

Hence P(1) is true for n = 1

Let P(n) is true for n = k

2.7^k+3.5^k-5 is divisible by 24 for all k \in N

\therefore 2.7^k+3.5^k-5 = 24 \lambda  … … … … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow 2.7^{k+1}+3.5^{k+1}-5 

= 7( 2.7^k) + 5( 3.5^k) - 5

= 7 ( 24 \lambda - 3.5^k + 5) + 5 ( 3.5^k) - 5

= 7 \times 24 \lambda - 7 ( 3.5^k) + 35 + 5 ( 3.5^k) - 5

= 7 \times 24 \lambda - 2 ( 3.5^k) + 30

= 7 \times 24 \lambda - 6 ( 5^k - 5)

We know that 5^k - 5 is divisible by 4 . Hence

= 7 \times 24 \lambda - 24 \mu

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

2.7^n+3.5^n-5 is divisible by 24 for all n \in N

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Question 27: 11^{n+2}+ 12^{2n+1} is divisible by 133 for all n \in N

Answer:

Let P(n) : 11^{n+2}+ 12^{2n+1} is divisible by 133 for all n \in N

For n = 1 , L.H.S = 11^3 + 12^3 = 3059 = 133 \times 23  which is divisible by 133

Hence P(1) is true for n = 1

Let P(n) is true for n = k

11^{k+2}+ 12^{2k+1} is divisible by 133 for all k \in N

\therefore 11^{k+2}+ 12^{2k+1} = 133 \lambda  … … … … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow 11^{(k+1)+2}+ 12^{2(k+1)+1}

= 11. 11^{k+2} + 144 .12^{2k+1}

= 11( 133 \lambda - 12^{2k+1}) + 144 . 12^{2k+1}

= 11 \times 133 \lambda - 11. 12^{2k+1} + 144 . 12^{2k+1}

= 11 \times 133 \lambda + 133. 12^{2k+1}

= 133 ( 11 \lambda + 12 ^{2k+1})

= 133 \mu where \mu = 11 \lambda + 12 ^{2k+1}  

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

2.7^n+3.5^n-5 is divisible by 24 for all n \in N

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Question 28: 1\times 1! + 2\times 2! +3\times 3! + \ldots + n\times n! = (n+1)! - 1 for all n \in N

Answer:

Let P(n) : 1\times 1! + 2\times 2! +3\times 3! + \ldots + n\times n! = (n+1)! - 1 for all n \in N

For n = 1 , L.H.S = 1\times 1!  = 1  R.H.S = 2! - 1  = 1 Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

1\times 1! + 2\times 2! +3\times 3! + \ldots + k\times k! = (k+1)! - 1 for all k \in N    … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow 1\times 1! + 2\times 2! +3\times 3! + \ldots + k\times k!  + (k+1)\times (k+1)! 

= (k+1)! - 1 + (k+1) \times (k+1)!

= (k+1)! [ 1 + (k+1)] - 1

= (k+1)! \times (k+2) - 1

= (k+2)! - 1

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

1\times 1! + 2\times 2! +3\times 3! + \ldots + n\times n! = (n+1)! - 1 for all n \in N

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Question 29: n^3 - 7n+3 is divisible by 3 for all n \in N

Answer:

Let P(n) : n^3 - 7n+3 is divisible by 3 for all n \in N

For n = 1 , L.H.S = 1 - 7 + 3 = -3  which is divisible by 3

Hence P(1) is true for n = 1

Let P(n) is true for n = k

k^3 - 7k+3 is divisible by 3 for all k \in N

\therefore k^3 - 7k+3 = 3 \lambda  … … … … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow (k+1)^3 - 7(k+1)+3

= k^3 + 1 + 3k^2 + 3k - 7 k - 7 + 3

= (k^3 - 7 k + 3) + ( 3k^2 + 3k - 6)

= 3 \lambda + 3 ( k^2 + k - 2)

= 3 \lambda + 3 \mu   where \mu = k^2 + k - 2

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

n^3 - 7n+3 is divisible by 3 for all n \in N

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Question 30: 1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1 for all n \in N

Answer:

Let P(n) : 1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1 for all n \in N

For n = 1 , L.H.S = 1+ 2^1 = 3  R.H.S = 2^2 - 1  = 3 Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

1 + 2 + 2^2 + \ldots + 2^k = 2^{k+1} - 1 for all k \in N   … … … … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow 1 + 2 + 2^2 + \ldots + 2^k + 2^{k+1} 

= 2^{k+1} - 1 + 2^{k+1}

= 2. 2^{k+1} - 1

= 2^{k+2}-1  

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1 for all n \in N

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Question 31: \text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{n-digits} =\frac{7}{81} (10^{n+1} - 9n-10) \text{ for all } n \in N

Answer:

Let P(n) : \text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{n-digits} =\frac{7}{81} (10^{n+1} - 9n-10) \text{ for all } n \in N

For n = 7 , L.H.S = 7  R.H.S = \frac{7}{81} ( 10^2 - 9 - 10) = 7 Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

\text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{k-digits} =\frac{7}{81} (10^{k+1} - 9k-10) \text{ for all } k \in N … … … … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{k-digits} + \underbrace{777 \ldots \ldots 7 }_{k+1-digits}

=\frac{7}{81} (10^{k+1} - 9k-10) + \underbrace{777 \ldots \ldots 7 }_{k+1-digits}

=\frac{7}{81} (10^{k+1} - 9k-10) + 7 ( \underbrace{111 \ldots \ldots 1 }_{k+1-digits} )

=\frac{7}{81} (10^{k+1} - 9k-10) + \frac{7}{9} ( \underbrace{999 \ldots \ldots 9 }_{k+1-digits} )

=\frac{7}{81} (10^{k+1} - 9k-10) + \frac{7}{9} ( 10^{k+1} - 1)

=\frac{7}{81} (10^{k+1} - 9k-10) + \frac{7}{81} ( 9.10^{k+1} - 9)

= \frac{7}{81} ( 10^{k+1} - 9k - 10 + 9.10^{k+1} - 9)

= \frac{7}{81} ( 10^{k+2} - 9 ( k+1) - 10)

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\text{ Prove that} \ 7 + 77 + 777+ \ldots + \underbrace{777 \ldots \ldots 7 }_{n-digits} =\frac{7}{81} (10^{n+1} - 9n-10) \text{ for all } n \in N

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Question 32: Prove that \frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210} n is a positive integer got all n \in N

Answer:

Let P(n) : \frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210} n is a positive integer got all n \in N

For n = 1 , L.H.S = \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210} = \frac{30+42+70+105-37}{210} = \frac{210}{210} = 1 which is positive.

Hence P(1) is true for n = 1

Let P(n) is true for n = k

\frac{k^7}{7} + \frac{k^5}{5} + \frac{k^3}{3} + \frac{k^2}{2} - \frac{37}{210} k is a positive integer got all k \in N … … … … … i)

Now we have to show that P(n) is true for n = k+1

\Rightarrow \frac{(k+1)^7}{7} + \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac{(k+1)^2}{2} - \frac{37}{210} (k+1)  

= \frac{1}{7} [ k^7 + 7k^6+ 21k^5 + 35k^4 + 35 k^3 + 21 k^2 + 7k + 1] + \frac{1}{5} [ k^5 + 5k^4+10k^3+10k^2+5k+1] + \frac{1}{3} [ k^3 + 3k^2 + 3k + 1] + \frac{1}{2} [k^2 + 2k + 1] - \frac{37}{210} (k+1)

= \Big[  \frac{k^7}{7} + \frac{k^5}{5} + \frac{k^3}{3} + \frac{k^2}{2} - \frac{37k}{210} \Big] + \Big[ k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k +\frac{1}{7} + k^4 + 2k^3 + 2k^2 + \frac{1}{5} + k^2 + k + \frac{1}{3}+ k + \frac{1}{2} - \frac{37}{210} \Big]

= \lambda + [ k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k + k^4 + 2k^3 + 2k^2 + k^2 + k + k ] + \Big[ \frac{1}{7} + \frac{1}{5} + \frac{1}{3} + \frac{1}{2} - \frac{37}{210}  \Big]

= \lambda + [k^6 + 3k^5 + 6k^4 + 7k^3 + 6k^2 + 3k] + 1

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210} n is a positive integer got all n \in N

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Question 33: Prove that \frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165} n is a positive integer got all n \in N

Answer:

Let P(n) : \frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165} n is a positive integer got all n \in N

For n = 1 ,

L.H.S = \frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165} = \frac{15+33+55+62}{165} = 1

1 is an integer. Hence P(1) is true for n = 1

Let P(n) is true for n = k

\frac{k^{11}}{11} + \frac{k^5}{5} + \frac{k^3}{3} + \frac{62}{165} k is a positive integer got all k \in N

Now we have to show that P(n) is true for n = k+1

\frac{(k+1)^{11}}{11} + \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac{62}{165} (k+1)

= \frac{1}{11} [ k^{11}+ 11k^{10}+55k^9 + 165k^8 +330k^7+ 462k^6+ 462k^5 + 330k^4 + 165k^3+ 55k^2+11k+1] + \frac{1}{5} [ k^5+5k^4+10k^3+10k^2+5k+1 ] + \frac{1}{3} [ k^3+3k^2+3k+1 ] + \frac{62}{165} [k+1]

= \Big[  \frac{k^{11}}{11} + \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{62k}{165} \Big] + k^{10} + 5k^9 + 15k^8 + 30 k^7 + 42k^6 + 42k^5 + 30k^4 + 15k^3 + 5k^2 +1 + \frac{1}{11} + k^4 + 2k^3 + 2k^2 + k + \frac{1}{5} + k^2 + k + \frac{1}{3} + \frac{62}{165}

= \Big[  \frac{k^{11}}{11} + \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{62k}{165} \Big] + k^{10} + 5k^9 + 15k^8 + 30k^7 + 42k^6 + 42k^5 + 31k^4 + 17k^3 + 8k^2 + 2k + 1

Which is an integer. \Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165} n is a positive integer got all n \in N

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Question 34: Prove that

\frac{1}{2} \tan \Big( \frac{x}{2} \Big) + \frac{1}{4} \tan \Big( \frac{x}{4} \Big) + \ldots + \frac{1}{2^n} \tan \Big( \frac{x}{2^n} \Big) = \frac{1}{2^n} \cot \Big( \frac{x}{2^n} \Big) - \cot x

for all n \in N and 0 < x < \frac{\pi}{2}

Answer:

Let P(n) :

\frac{1}{2} \tan \Big( \frac{x}{2} \Big) + \frac{1}{4} \tan \Big( \frac{x}{4} \Big) + \ldots + \frac{1}{2^n} \tan \Big( \frac{x}{2^n} \Big) = \frac{1}{2^n} \cot \Big( \frac{x}{2^n} \Big) - \cot x

for all n \in N and 0 < x < \frac{\pi}{2}

For n = 1 , L.H.S = \frac{1}{2} \tan \frac{x}{2}  

R.H.S = \frac{1}{2} \cot \frac{x}{2} -\cot x =  \frac{1}{2 \tan \frac{x}{2}} - \frac{1}{\tan x}

= \frac{1}{2 \tan \frac{x}{2}} - \frac{1}{\frac{2 \tan \frac{x}{2}}{1 - \tan^2 \frac{x}{2}}}

= \frac{1-1+\tan^2 \frac{x}{2}}{2 \tan \frac{x}{2}} = \frac{1}{2} \tan \frac{x}{2}

Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

\frac{1}{2} \tan \Big( \frac{x}{2} \Big) + \frac{1}{4} \tan \Big( \frac{x}{4} \Big) + \ldots + \frac{1}{2^k} \tan \Big( \frac{x}{2^k} \Big) = \frac{1}{2^k} \cot \Big( \frac{x}{2^k} \Big) - \cot x  … i)

Now we have to show that P(n) is true for n = k+1

\frac{1}{2}  \tan \big(  \frac{x}{2}  \big) +  \frac{1}{4}  \tan \big(  \frac{x}{4}  \big) + \ldots + \frac{1}{2^k}  \tan \big(  \frac{x}{2^k}  \big) + \frac{1}{2^{k+1}}  \tan \big(  \frac{x}{2^{k+1}}  \big)

= \frac{1}{2^k}  \cot \big(  \frac{x}{2^k}  \big) - \cot x + \frac{1}{2^{k+1}}  \tan \big(  \frac{x}{2^{k+1}}  \big)

= \frac{1}{2^k}  \Big[  \cot \big(  \frac{x}{2^k}  \big) + \frac{1}{2}  \tan \big(  \frac{x}{2^{k+1}}  \big)  \Big] - \cot x

= \frac{1}{2^k}  \Big[  \frac{1}{\tan \big(  \frac{x}{2^k}  \big)} + \frac{1}{2}  \tan \big(  \frac{x}{2^{k+1}}  \big)  \Big] - \cot x

= \frac{1}{2^k}  \Big[  \frac{1}{\tan \big(  \frac{x}{2^{k+1}} + \frac{x}{2^{k+1}}  \big)} + \frac{1}{2}  \tan \big(  \frac{x}{2^{k+1}}  \big)  \Big] - \cot x

= \frac{1}{2^k}  \Big[ \frac{1 - \tan^2 \frac{x}{2^{k+1}}}{2 \tan \frac{x}{2^{k+1}}}  + \frac{1}{2}  \tan \big(  \frac{x}{2^{k+1}}  \big)  \Big] - \cot x

= \frac{1}{2^{k+1} } \Big[  \frac{1 - \tan^2 \frac{x}{ 2^{k+1} } + \tan^2 \frac{x}{ 2^{k+1} }}{\tan^2 \frac{x}{ 2^{k+1} } }    \Big] - \cot x

= \frac{1}{2^{k+1} } \cot^2 \frac{x}{ 2^{k+1} } - \cot x

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{1}{2} \tan \Big( \frac{x}{2} \Big) + \frac{1}{4} \tan \Big( \frac{x}{4} \Big) + \ldots + \frac{1}{2^n} \tan \Big( \frac{x}{2^n} \Big) = \frac{1}{2^n} \cot \Big( \frac{x}{2^n} \Big) - \cot x

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Question 35: Prove that

\Big( 1 - \frac{1}{2^2} \Big) \Big( 1 - \frac{1}{3^2} \Big) \Big( 1 - \frac{1}{4^2} \Big) \ldots \Big( 1 - \frac{1}{n^2} \Big) = \frac{n+1}{2n} for all natural numbers, n \geq 2

Answer:

Let P(n) : \Big( 1 - \frac{1}{2^2} \Big) \Big( 1 - \frac{1}{3^2} \Big) \Big( 1 - \frac{1}{4^2} \Big) \ldots \Big( 1 - \frac{1}{n^2} \Big) = \frac{n+1}{2n} for all natural numbers, n \geq 2

For n = 2 , L.H.S = \big(  1 - \frac{1}{2^2} \big) = \frac{3}{4}   R.H.S = \frac{2+1}{2 \cdot 2} = \frac{3}{4} Therefore LHS = RHS

Hence P(2) is true for n = 2

Let P(n) is true for n = k

\Big( 1 - \frac{1}{2^2} \Big) \Big( 1 - \frac{1}{3^2} \Big) \Big( 1 - \frac{1}{4^2} \Big) \ldots \Big( 1 - \frac{1}{k^2} \Big) = \frac{k+1}{2k} for all natural numbers, k \geq 2  … … i)

Now we have to show that P(n) is true for n = k+1

\Big( 1 - \frac{1}{2^2} \Big) \Big( 1 - \frac{1}{3^2} \Big) \Big( 1 - \frac{1}{4^2} \Big) \ldots \Big( 1 - \frac{1}{k^2} \Big) \Big( 1 - \frac{1}{(k+1)^2} \Big)

= \Big( \frac{k+1}{2k} \Big) \Big( 1 - \frac{1}{(k+1)^2} \Big)

= \frac{k+1}{2k} \Big( \frac{k^2+1 + 2k - 1}{(k+1)^2} \Big)

= \frac{k(k+2)}{2k ( k+1)}

= \frac{k+2}{2(k+1)}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\Big( 1 - \frac{1}{2^2} \Big) \Big( 1 - \frac{1}{3^2} \Big) \Big( 1 - \frac{1}{4^2} \Big) \ldots \Big( 1 - \frac{1}{n^2} \Big) = \frac{n+1}{2n} for all natural numbers, n \geq 2

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Question 36: Prove that \frac{ (2n)!}{2^{2n}(n!)^2} \leq \frac{1}{\sqrt{3n+1}} for all n \in N

Answer:

Let P(n) : \frac{ (2n)!}{2^{2n}(n!)^2} \leq \frac{1}{\sqrt{3n+1}} for all n \in N

For n = 1 , = \frac{2!}{2^2 \cdot 1} = \frac{1}{2} \leq \frac{1}{\sqrt{3+1}}

Hence P(1) is true for n = 1

Let P(n) is true for n = k

\frac{ (2k)!}{2^{2k}(k!)^2} \leq \frac{1}{\sqrt{3k+1}} for all n \in N … … … … … i)

Now we have to show that P(n) is true for n = k+1

\frac{ [2(k+1)]!}{2^{2(k+1)}[(k+1)!]^2}

= \frac{(2k+2)(2k+1)(2k)!}{2^{2k} \cdot 2^2 \cdot ( k+1) k! (k+1) k!}

= \frac{(2k+1) (2k)!}{2^{2k} \cdot 2 k! (k+1) k!}

= \frac{(2k)!}{2^{2k} (k!)^2} \Big[ \frac{2k+1}{2 (k+1)} \Big]

\leq \frac{1}{\sqrt{3k+1}} \Big[  \frac{2k+1}{2 (k+1)} \Big]

Since 2k+1 < 2k+2 \Rightarrow \frac{2k+1}{2(k+1)} < \frac{2k+2}{2(k+1)} = 1

\leq \frac{1}{\sqrt{3k+1}}

Since 3k+1 < 3k+4

\leq \frac{1}{\sqrt{3k+4}}

\leq \frac{1}{\sqrt{3(k+1)+1}}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{ (2n)!}{2^{2n}(n!)^2} \leq \frac{1}{\sqrt{3n+1}} for all n \in N

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Question 37: 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots + \frac{1}{n^2} < 2 - \frac{1}{n} for all n \geq 2, n \in N

Answer:

Let P(n) : 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots + \frac{1}{n^2} < 2 - \frac{1}{n} for all n \geq 2, n \in N

For n = 2 , \frac{1}{2^2} = \frac{1}{4} < 2 - \frac{1}{2}

Hence P(2) is true for n = 2

Let P(n) is true for n = k

1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots + \frac{1}{k^2} < 2 - \frac{1}{k} for all k \geq 2, n \in N … … … … … i)

Now we have to show that P(n) is true for n = k+1

1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots + \frac{1}{k^2} + \frac{1}{(k+1)^2}

< 2 - \frac{1}{k} + \frac{1}{(k+1)^2}

< 2 - \Big[  \frac{(k+1)^2 - k}{k(k+1)^2} \Big]

< 2 - \Big[  \frac{(k^2 + 1 + 2k - k}{k(k+1)^2} \Big]

< 2 - \Big[  \frac{(k^2 + k+1}{k(k+1)^2} \Big]

< 2 - \frac{k^2 + k}{k(k+1)^2}

< 2 - \frac{1}{k+1}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots + \frac{1}{n^2} < 2 - \frac{1}{n} for all n \geq 2, n \in N

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Question 38: Prove that x^{2n-1} + y^{2n-1} is divisible by x+y for all n \in N

Answer:

Let P(n) : x^{2n-1} + y^{2n-1} is divisible by x+y for all n \in N

For n = 1 , x^{2-1}+y^{2-1} = x+y which is divisible by x+y

Hence P(1) is true for n = 1

Let P(n) is true for n = k

x^{2k-1} + y^{2k-1} is divisible by x+y for all k \in N

\Rightarrow x^{2k-1} + y^{2k-1} = (x+y) \mu    … … … … i)

Now we have to show that P(n) is true for n = k+1

x^{2(k+1) - 1} - y^{2(k+1) - 1}

= x^{2k+1} - y^{2k+1}

= x^{2k -1 +2} - y^{2k- 1+2}

= x^2 \cdot x^{2k-1} - y^2 \cdot y^{2k-1}

= x^2 [ (x+y) \mu + y^{2k-1} ] - y^2 \cdot y^{2k-1}

= x^2 (x+y) \mu + x^2 y^{2k-1} - y^2 \cdot y^{2k-1}

= x^2 (x+y) \mu + (x^2 - y^2) y^{2k-1}

= x^2 (x+y) \mu + (x+y)(x-y) y^{2k-1}

= (x+y) [ x^2 \mu + (x-y) y^{2k-1} ]

= (x+y) \lambda where \lambda = [ x^2 \mu + (x-y) y^{2k-1} ]

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

x^{2n-1} + y^{2n-1} is divisible by x+y for all n \in N

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Question 39: Prove that \sin x + \sin 3x + \ldots + \sin (2n-1) x = \frac{\sin^2 nx}{\sin x} for all n \in N

Answer:

Let P(n) : \sin x + \sin 3x + \ldots + \sin (2n-1) x = \frac{\sin^2 nx}{\sin x} for all n \in N

For n = 1 , LHS = \sin x   RHS = \frac{\sin^2 x}{\sin x} = \sin x Therefore LHS = RHS Hence latex P(1) $ is true for n = 1

Let P(n) is true for n = k

\sin x + \sin 3x + \ldots + \sin (2k-1) x = \frac{\sin^2 kx}{\sin x} for all k \in N    … … … i)

Now we have to show that P(n) is true for n = k+1

\sin x + \sin 3x + \ldots + \sin (2k-1) x + \sin [2(k+1) - 1]x

= \frac{\sin^2 kx}{\sin x} + \sin [2(k+1) - 1]x

= \frac{\sin^2 kx}{\sin x} + \sin (2k+1) x

= \frac{\sin^2 kx + \sin x \cdot \sin (2k+1) x}{\sin x}

= \frac{\sin^2 kx + \{  \cos [ (2k+1)x - x] - \cos [ (2k+1)x + x ] \}  }{2\sin x}

= \frac{\sin^2 kx + \cos 2kx - \cos [2x(k+1)]  }{2\sin x}

= \frac{1 - \cos 2kx + \cos 2kx  - \cos 2x(k+1)  }{2\sin x}

= \frac{1- \cos 2x(k+1)  }{2\sin x}

= \frac{2 \sin^2 x (k+1)  }{2\sin x}

= \frac{ \sin^2 x (k+1)  }{\sin x}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\sin x + \sin 3x + \ldots + \sin (2n-1) x = \frac{\sin^2 nx}{\sin x} for all n \in N

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Question 40: Prove that \cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (n-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{n-1}{2} \big) \beta \big\} \sin \big( \frac{n \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }

Answer:

Let P(n) :

\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (n-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{n-1}{2} \big) \beta \big\} \sin \big( \frac{n \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }

For n = 1 , LHS = \cos ( \alpha + (1-1) \beta) = \cos \alpha

RHS = \frac{\cos \big\{ \alpha + \big( \frac{1-1}{2} \big) \beta \big\} \sin \big( \frac{\beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) } = \cos \alpha

Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (k-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{k-1}{2} \big) \beta \big\} \sin \big( \frac{k \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }   … i)

Now we have to show that P(n) is true for n = k+1

\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (k-1) \beta) + \cos ( \alpha + k \beta)

= \frac{\cos \big\{ \alpha + \big( \frac{k-1}{2} \big) \beta \big\} \sin \big( \frac{k \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) } + \cos ( \alpha + k \beta)

= \frac{\sin \frac{\beta}{2} \cos ( \alpha + k \beta) + \cos \big\{ \alpha + \big( \frac{k-1}{2} \big) \beta \big\} \sin \big( \frac{k \beta}{2} \big)}{\sin \big( \frac{\beta}{2} \big)}

= \frac{\frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k+1}{2} \big) \beta  \} - \sin \{ \alpha + \big( \frac{2k-1}{2} \big) \beta \}    \Big] +         \cos \{ \alpha + \big( \frac{k-1}{2} \big) \beta \} \sin \frac{k\beta}{2}  }{\sin \big( \frac{\beta}{2} \big)}

= \frac{\frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k+1}{2} \big) \beta  \} - \sin \{ \alpha + \big( \frac{2k-1}{2} \big) \beta \}    \Big] + \frac{1}{2} \Big[  \sin \{ \alpha + \big( \frac{2k-1}{2} \big) \beta \}  + \sin \big( - \alpha + \frac{\beta}{2}  \big)   \Big] }{\sin \big( \frac{\beta}{2} \big)}

= \frac{\frac{1}{2} \Big[ \sin \{ \alpha + \big( \frac{2k+1}{2} \big) \beta  \}   + \sin \big( - \alpha + \frac{\beta}{2}  \big)   \Big] }{\sin \big( \frac{\beta}{2} \big)}

= \frac{\frac{1}{2} \Big[ \sin \alpha \cos \big( \frac{2k+1}{2} \big) \beta + \cos \alpha \sin \big( \frac{2k+1}{2} \big) \beta  + \sin \frac{\beta}{2} \cos \alpha -  \cos \frac{\beta}{2} \sin \alpha \Big] }{\sin \big( \frac{\beta}{2} \big)}

= \frac{\frac{1}{2} \Big[ \sin \alpha \Big\{ \cos \big( \frac{2k+1}{2} \big) \beta - \cos \frac{\beta}{2} \Big\} + \cos \alpha \Big\{ \sin \big( \frac{2k+1}{2} \big) \beta  + \sin \frac{\beta}{2} \Big\}  \Big] }{\sin \big( \frac{\beta}{2} \big)}

= \frac{\frac{1}{2} \Big[ -2\sin \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\} + 2\cos \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\}  \Big] }{\sin \big( \frac{\beta}{2} \big)}

= \frac{ \Big[ -2\sin \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\} + \cos \alpha \Big\{ \sin \big( \frac{k+1}{2} \big) \beta \cdot \sin \frac{k\beta}{2} \Big\}  \Big] }{\sin \big( \frac{\beta}{2} \big)}

= \frac{ \sin \big( \frac{k+1}{2} \big) \beta  \Big[ \cos \alpha  \cdot \cos \frac{k\beta}{2} -  \sin \alpha \cdot \sin \frac{k\beta}{2}   \Big]  }{\sin \big( \frac{\beta}{2} \big)}

= \frac{ \sin \big( \frac{k+1}{2} \big) \beta  \cos \big( \alpha + \frac{k\beta}{2} \big)  }{\sin \big( \frac{\beta}{2} \big)}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\cos \alpha + \cos ( \alpha + \beta) + \cos ( \alpha + 2\beta) + \ldots + \cos ( \alpha + (n-1) \beta) = \frac{\cos \big\{ \alpha + \big( \frac{n-1}{2} \big) \beta \big\} \sin \big( \frac{n \beta}{2} \big) }{\sin \big( \frac{\beta}{2} \big) }

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Question 41: Prove that \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} > \frac{13}{24} , for all natural numbers n > 1

Answer:

Let P(n) : \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} > \frac{13}{24} , for all natural numbers n > 1

For n = 2 , L.H.S = \frac{1}{2+1} + \frac{1}{2 \times 2}= \frac{1}{3} + \frac{1}{4} = \frac{14}{24} > \frac{13}{24} =  R.H.S

Hence P(2) is true for n = 2

Let P(n) is true for n = k

\frac{1}{k+1} + \frac{1}{k+2} + \ldots + \frac{1}{2k} > \frac{13}{24} , for all natural numbers k > 1 … … … … … i)

Now we have to show that P(n) is true for n = k+1

\frac{1}{k+2} + \ldots + \frac{1}{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}

= - \frac{1}{k+1} + \frac{1}{k+1}\frac{1}{k+2} + \ldots + \frac{1}{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}

> \frac{13}{24} - \frac{1}{k+1} + \frac{1}{2k+1} + \frac{1}{2k+2}

> \frac{13}{24} - \frac{1}{2(k+1)} + \frac{1}{2k+1}

> \frac{13}{24} + \frac{2(k+1)-(2k+1)}{2(k+1)(2k+1)}

> \frac{13}{24} + \frac{2k+2-2k-1}{2(k+1)(2k+1)}

> \frac{13}{24} + \frac{1}{2(k+1)(2k+1)}

> \frac{13}{24}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} > \frac{13}{24} , for all natural numbers n > 1

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Question 42: Give

a_1 = \frac{1}{2} \Big( a_0 + \frac{A}{a_0} \Big), a_2 = \frac{1}{2} \Big( a_1 + \frac{A}{a_1} \Big) and a_{n+1} = \frac{1}{2} \Big( a_n + \frac{A}{a_n} \Big)

for n \geq 2 , where a > 0 , A > 0

Prove that \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \Big( \frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}} \Big)^{2^{n-1}}

Answer:

Let P(n) : \frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \Big( \frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}} \Big)^{2^{n-1}}

For n = 1     LHS = \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}

RHS = \Big( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \Big)^{2-1} = \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}}

Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

\frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} = \Big( \frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}} \Big)^{2^{k-1}} is true for k \leq 2 … … … … … i)

Now we have to show that P(n) is true for n = k+1

LHS = \frac{a_{k+1} - \sqrt{A}}{a_{k+1} + \sqrt{A}}

= \frac{ \frac{1}{2} \big( a_k + \frac{A}{a_k} \big) - \sqrt{A}}{\frac{1}{2} \big( a_k + \frac{A}{a_k} \big) + \sqrt{A}}

= \frac{\frac{1}{2} \big( \frac{{a_k}^2+A - 2 a_k \sqrt{A}}{a_k}  \big) }{\frac{1}{2} \big( \frac{{a_k}^2+A + 2 a_k \sqrt{A}}{a_k}  \big)}

= \frac{{a_k}^2+A - 2a_k \sqrt{A}}{{a_k}^2+A + 2a_k \sqrt{A}}

= \frac{(a_k - \sqrt{A})^2}{(a_k + \sqrt{A})^2}

= \Big( \frac{a_k - \sqrt{A}}{a_k + \sqrt{A}} \Big)^2

= \Big[ \Big( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \Big)^{2^{k-1}} \Big]^2

= \Big( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \Big)^{2^k}

RHS = \Big( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \Big)^{2^{k+1-1}}

= \Big( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \Big)^{2^k}

Therefore LHS = RHS

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \Big( \frac{a_1 - \sqrt{A}}{a_1+\sqrt{A}} \Big)^{2^{n-1}}

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Question 43: Let P(n) be the statement: 2^n \geq 3n . If P(r) is true, show that P(r+1) is true. Do you conclude that P(n) is true for all n \in N ?

Answer:

Let P(n) : 2^n \geq 3n for all n \in N

For n = 1 ,  LHS = 2^1 = 2     RHS = 3 \times 1 = 3

Therefore LHS < RHS

Hence P(1) is NOT true for n = 1

Therefore P(n) is NOT true for all n \in N

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Question 44: Show that by the principle of mathematical induction that the sum S_n of the n terms of the series 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots is given by

S_n = \Bigg\{ \begin{array}{ll} \frac{n(n+1)^2}{2}, \text{if} \ n \ \text{is even} \\   \frac{n^2(n+1)}{2} , \text{if} \ n \ \text{is odd} \end{array}

Answer:

Let P(n) :

S_n of the n terms of the series 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots is given by

S_n = \Bigg\{ \begin{array}{ll} \frac{n(n+1)^2}{2}, \text{if} \ n \ \text{is even} \\   \frac{n^2(n+1)}{2} , \text{if} \ n \ \text{is odd} \end{array}

For n = 2 ,

L.H.S S_2= 1^2 + 2 \times 2^2 = 9

R.H.S S_2= \frac{2(3)^2}{2} = 9

Therefore  LHS = RHS

Hence P(n) is true for n = 2

For n = 3 ,

L.H.S S_3= 1^2 + 2 \times 2^2 + 3^2 = 18

R.H.S S_3= \frac{3^2(4)}{2} = 18

Therefore  LHS = RHS

Hence P(n) is true for n = 3

Let P(n) is true for n = k

S_k of the k terms of the series 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots is given by

S_k = \Bigg\{ \begin{array}{ll} \frac{k(k+1)^2}{2}, \text{if} \ k \ \text{is even} \\   \frac{k^2(k+1)}{2} , \text{if} \ k \ \text{is odd} \end{array}   … … … … … i)

Now we have to show that P(n) is true for n = k+1

If  k is even,  (k+1) is odd

\therefore S_{k+1} = 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots  + 2k^2 + ( k+1)^2

S_{k+1} = S_k + ( k+1)^2

\Rightarrow S_{k+1} = \frac{k(k+1)^2}{2} + ( k+1)^2 = (k+1)^2 \big( \frac{k+2}{2} \big)

If  k is odd,  (k+1) is even

\therefore S_{k+1} = 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots  + 2k^2 + 2( k+1)^2

S_{k+1} = S_k + 2( k+1)^2

\Rightarrow S_{k+1} = \frac{k^2(k+1)}{2} + 2( k+1)^2 = (k+1) \big( \frac{(k+2)^2}{2} \big)

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

S_n of the n terms of the series 1^2+ 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + \ldots is given by

S_n = \Bigg\{ \begin{array}{ll} \frac{n(n+1)^2}{2}, \text{if} \ n \ \text{is even} \\   \frac{n^2(n+1)}{2} , \text{if} \ n \ \text{is odd} \end{array}

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Question 45: Prove that the number of subsets of a set containing n distinct elements is 2^n for all n \in N

Answer:

Let P(n) : The number of subsets of a set containing n distinct elements is 2^n for all n \in N

For n = 1  LHS = number of subsets of a set containing only 1 element = 2 ( \phi and the set itself  )   RHS = 2^1= 2 .

Hence P(1) is true for n = 1

Let P(n) is true for n = k

The number of subsets of a set containing k distinct elements is 2^k for all k \in N

Now we have to show that P(n) is true for n = k+1

Let A = \{1 ,2, 3, \ldots, k, k+1 \}

Now A = \{1 ,2, 3, \ldots, k \} \cup \{m+1 \}

Using i) we can say that \{1 ,2, 3, \ldots, k \} has 2^k subset and \{ m+1 \} as 2^1 subset.

\Rightarrow A has 2^k + 2 = s^{k+1} subsets.

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction The number of subsets of a set containing n distinct elements is 2^n for all n \in N

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Question 46: A sequence a_1, a_2, a_3, \ldots is defined by letting a_1 = 3 and a_k = 7 a_{k-1} for all natural numbers k \geq 2 . Show that a_n = 3 \cdot 7^{n-1} for all n \in N .

Answer:

Given a sequence a_1, a_2, a_3, \ldots is defined by letting a_1 = 3 and a_k = 7 a_{k-1} for all natural numbers k \geq 2

Let P(n) : a_n = 3 \cdot 7^{n-1} for all n \in N

For n = 1 , L.H.S = a_1 = 3 \cdot 7^{1-1} = 3 \cdot 1 = 1

Hence P(1) is true for n = 1

Let P(n) is true for n = k

a_k = 3 \cdot 7^{k-1} for all k \in N … … i)

Now we have to show that P(n) is true for n = k+1

x_{k+1} = \frac{x_{k+1-1}}{k}

a_{k+1} = 7 a_k

= 7 \cdot 3  \cdot 7^{k-1}

= 3 \cdot 7^{k-1+1}

= 3 \cdot 7^{(k+1)-1}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction a_n = 3 \cdot 7^{n-1} for all n \in N

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Question 47: A sequence x_1, x_2, x_3, \ldots is defined by letting x_1 = 2 and x_k = \frac{x_{k-1}}{n} for all natural numbers k, k \geq 2 . Show that x_n = \frac{2}{n!} for all n \in N .

Answer:

Given a sequence x_1, x_2, x_3, \ldots is defined by letting x_1 = 2 and x_k = \frac{x_{k-1}}{n} for all natural numbers k, k \geq 2 .

Let P(n) : x_n = \frac{2}{n!} for all n \in N .

For n = 1 , L.H.S = x_1 = \frac{2}{1!} = 2

Hence P(1) is true for n = 1

Let P(n) is true for n = k

x_k = \frac{2}{k!} for all k \in N … … i)

Now we have to show that P(n) is true for n = k+1

x_{k+1} = \frac{x_{k+1-1}}{k}

= \frac{x_k}{k}

= \frac{2}{k \times k!}

= \frac{2}{(k+1)!}

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction x_n = \frac{2}{n!} for all n \in N

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Question 48: A sequence x_0, x_1, x_2, x_3, \ldots is defined by letting x_0 = 5 and x_k = 4+x_{k-1} for all natural numbers k . Show that x_n = 5+4n for all n \in N using mathematical induction.

Answer:

Given A sequence x_0, x_1, x_2, x_3, \ldots is defined by letting x_0 = 5 and x_k = 4+x_{k-1} for all natural numbers k .

Let P(n) : x_n = 5+4n for all n \in N

For n = 0 , L.H.S = x_0 = 5 + 4 \times 0 = 5

Hence P(0) is true for n = 0

Let P(n) is true for n = k

x_k = 5+4k for all n \in N  … … i)

Now we have to show that P(n) is true for n = k+1

x_{k+1} = 4 + x_{k+1-1}

= 4 + x_{k}

= 4 + 5 + 4k

= 5 + 4 ( k+1)

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

x_n = 5+4n for all n \in N

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Question 49: Using the principle of mathematical induction prove that

\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} for all natural numbers n \geq 2 .

Answer:

Let P(n) : \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} for all natural numbers n \geq 2

For n = 2 , L.H.S = \sqrt{2}  = 1.41  R.H.S = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}}   = 1 + 0.707 = 1.707 Therefore LHS < RHS

Hence P(2) is true for n = 2

Let P(n) is true for n = k

\sqrt{k} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{k}} for all natural numbers k \geq 2  … … i)

Now we have to show that P(n) is true for n = k+1

LHS = \sqrt{k+1}

RHS = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}}

> \sqrt{k} + \frac{1}{\sqrt{k+1}}

Since \sqrt{k+1} > \sqrt{k}

\Rightarrow \frac{\sqrt{k}}{\sqrt{k+1}} < 1

\Rightarrow \frac{1}{\sqrt{k+1}} < \sqrt{k}

\Rightarrow \frac{k+1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+1}} < \sqrt{k}

\Rightarrow \sqrt{k+1} - \frac{1}{\sqrt{k+1}} < \sqrt{k}

\Rightarrow   \sqrt{k} + \frac{1}{\sqrt{k+1}} > \sqrt{k+1}

Therefore LHS < RHS

\Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} for all natural numbers n \geq 2

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Question 50: The distributive law from algebra states that for all real numbers c, a_1 and a_2 we have c(a_1+a_2) = ca_1 + ca_2 . Use this law and mathematical induction to prove that, for all natural numbers, n \geq 2 , if c, a_1, a_2, a_3, \ldots , a_n are any real numbers, then

c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n

Answer:

For all real numbers c, a_1 , and a_2, c( a_1 + a_2) = ca_1 + ca_2

To prove that, for all natural numbers, n \geq 2 , if c, a_1, a_2, a_3, \ldots , a_n are any real numbers, then c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n

Let P(n) : c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n for all natural numbers n \geq 2 , and c, a_1, a_2, \ldots , a_n \in R

For n = 2 , LHS = c( a_1+a_2) , RHS = ca_1 +ca_2  \Rightarrow LHS = RHS

Hence P(n) is true for n=2

Let P(n) is true for n = k therefore c(a_1+a_2+ \ldots + a_k) = ca_1 + ca_2 + \ldots + ca_k for all natural numbers n \geq 2 , and c, a_1, a_2, \ldots , a_k \in R

For n = k+1

c[a_1+a_2+ \ldots + a_k + a_{k+1}]

= c (a_1+a_2+ \ldots + a_k) + ca_{k+1}

= ca_1 + ca_2 + \ldots + ca_k + ca_{k+1}

Therefore P(n) is true for n = k+1

Hence by the principle of mathematical induction

c(a_1+a_2+ \ldots + a_n) = ca_1 + ca_2 + \ldots + ca_n for all natural numbers n \geq 2 , and c, a_1, a_2, \ldots , a_n \in R

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