Prove the following by the principle of mathematical induction:

Question 1: 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} i.e., the sum of the first n natural numbers is \frac{n(n+1)}{2} .

Answer:

Let P(n) : 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}

For n = 1 , L.H.S = 1 R.H.S = \frac{1(1+1)}{2} = 1

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2} … … … … … i)

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

1 + 2 + 3 + \ldots + k +(k+1)

Now substituting the from equation i)

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

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

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

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

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

Hence by the principle of mathematical induction

1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} is true for all n \in N

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Question 2: 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}

Answer:

Let P(n) : 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}

For n = 1 , L.H.S = 1 R.H.S = \frac{1(1+1)(2+1)}{6} = 1

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6} … … … … … i)

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

1^2 + 2^2 + 3^2 + \ldots + k^2 + (k+1)^2

Now substituting the from equation i)

= \frac{k(k+1)(2k+1)}{6} + (k+1)^2

= \frac{k+1}{6} [ k(2k+1) + 6(k+1) ]

= \frac{k+1}{6} [ 2k^2 + k + 6k + 6]

= \frac{(k+1)(2k+3)(k+2)}{6}

= \frac{(k+1)(k+1+1)(2(k+1)+1)}{6}

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

Hence by the principle of mathematical induction

1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} is true for all n \in N

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Question 3: 1+ 3 + 3^2 + \ldots + 3^{n-1} = \frac{3^n -1}{2}

Answer:

Let P(n) : 1+ 3 + 3^2 + \ldots + 3^{n-1} = \frac{3^n -1}{2}

For n = 1 , L.H.S = 1 R.H.S = \frac{3-1}{2} = 1

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1+ 3 + 3^2 + \ldots + 3^{k-1} = \frac{3^k -1}{2} … … … … … i)

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

1+ 3 + 3^2 + \ldots + 3^{k-1} + 3^k

Now substituting the from equation i)

= \frac{3^k -1}{2} + 3^k

= \frac{3^k - 1 + 2.3^k}{2}

= \frac{3.3^k - 1}{2}

= \frac{3^{k+1}-1}{2}

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

Hence by the principle of mathematical induction

1+ 3 + 3^2 + \ldots + 3^{n-1} = \frac{3^n -1}{2} is true for all n \in N

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Question 4: \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}

Answer:

Let P(n) : \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}

For n = 1 , L.H.S = \frac{1}{1.2} R.H.S = \frac{1}{2} = 1

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{k(k+1)} = \frac{k}{k+1} … … … … … i)

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

\Rightarrow \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}

Now substituting the from equation i)

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

= \frac{k}{k+1} \Big[ k+ \frac{1}{k+2} \Big]

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

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

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

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

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

Hence by the principle of mathematical induction

\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1} is true for all n \in N

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Question 5: 1+ 3+ 5 + \ldots + (2n-1) = n^2 i.e., the sum of firs n odd numbers is n^2

Answer:

Let P(n) : 1+ 3+ 5 + \ldots + (2n-1) = n^2

For n = 1 , L.H.S = 1 R.H.S = 1

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1+ 3+ 5 + \ldots + (2k-1) = k^2 … … … … … i)

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

1+ 3+ 5 + \ldots + (2k-1) + (2k+1)

Now substituting the from equation i)

= k^2 + 2k+1

=(k+1)^2

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

Hence by the principle of mathematical induction

1+ 3+ 5 + \ldots + (2n-1) = n^2 is true for all n \in N

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Question 6: \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}

Answer:

Let P(n) : \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}

For n = 1 , L.H.S = \frac{1}{10} R.H.S = \frac{1}{10}

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \ldots + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4} … … … … … i)

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

\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \ldots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3k+2)(3k+5)}

= \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)}

= \frac{1}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)}

= \frac{1}{3k+2} \Big[ \frac{k}{2} + \frac{1}{3k+5} \Big]

= \frac{1}{(3k+2)} \frac{3k^2+5k+2}{2(3k+5)}

= \frac{1}{(3k+2)} \frac{(3k+2)(k+1)}{2(3k+5)}

= \frac{k+1}{2(3k+5)}

= \frac{k+1}{6(k+1)+4}

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

Hence by the principle of mathematical induction

\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} is true for all n \in N

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Question 7: \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}

Answer:

Let P(n) : \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}

For n = 1 , L.H.S = \frac{1}{4} R.H.S = \frac{1}{4}

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots + \frac{1}{(3k-2)(3k+1)} = \frac{k}{3k+1} … … … … … i)

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

\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots + \frac{1}{(3k-2)(3k+1)} + \frac{1}{[3(k+1)-2][3(k+1)+1]}

= \frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)}

= \frac{1}{(3k+1)} \Big[ k + \frac{1}{3k+4} \Big]

= \frac{1}{(3k+1)} \frac{3k^2+4k+1}{(3k+4)}

= \frac{1}{(3k+1)} \frac{(3k+1)(k+1)}{(3k+4)}

= \frac{(k+1)}{(3k+4)}

= \frac{(k+1)}{(3(k+1)+1)}

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

Hence by the principle of mathematical induction

\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1} is true for all n \in N

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Question 8: \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}

Answer:

Let P(n) : \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}

For n = 1 , L.H.S = \frac{1}{15} R.H.S = \frac{1}{15}

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + \ldots + \frac{1}{(2k+1)(2k+3)} = \frac{k}{3(2k+3)} … … … … … i)

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

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Question 9: \frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + \ldots + \frac{1}{(4n-1)(4n+3)} = \frac{n}{3(4n+3)}

Answer:

Let P(n) : \frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + \ldots + \frac{1}{(4n-1)(4n+3)} = \frac{n}{3(4n+3)}

For n = 1 , L.H.S = \frac{1}{21} R.H.S = \frac{1}{21}

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow \frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + \ldots + \frac{1}{(4k-1)(4k+3)} = \frac{k}{3(4k+3)} … … … … … i)

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

\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + \ldots + \frac{1}{(4k-1)(4k+3)} + \frac{1}{[4(k+1)-1][4(k+1)+3]}

= \frac{k}{3(4k+3)} + \frac{1}{(4k+3)(4k+7)}

= \frac{1}{(4k+3)} \Big[ \frac{k}{3} + \frac{1}{4k+7} \Big]

= \frac{1}{(4k+3)} \Big[ \frac{4k^2+7k+3}{3(4k+7)} \Big]

= \frac{1}{(4k+3)} \Big[ \frac{(4k+3)(k+1)}{3(4k+7)} \Big]

= \frac{k+1}{3(4k+7)}

= \frac{k+1}{3[4(k+1)+3]}

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

Hence by the principle of mathematical induction

\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + \ldots + \frac{1}{(4n-1)(4n+3)} = \frac{n}{3(4n+3)} is true for all n \in N

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Question 10: 1.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = ( n-1)2^{n+1} + 2

Answer:

Let P(n) : 1.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = ( n-1)2^{n+1} + 2

For n = 1 , L.H.S = 2 R.H.S = 2

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1.2 + 2.2^2 + 3.2^3 + \ldots + k.2^k = ( k-1)2^{k+1} + 2 … … … … … i)

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

1.2 + 2.2^2 + 3.2^3 + \ldots + k.2^k + (k+1)2^{k+1}

= [(k-1)2^{k+1} + 2 ] + (k+1)2^{k+1}

= k2^{k+1}- 2^{k+1}+2 + k 2^{k+1}+ 2^{k+1}

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

= [(k+1)-1]2^{(k+1)+1} + 2

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

Hence by the principle of mathematical induction

1.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = ( n-1)2^{n+1} + 2 is true for all n \in N

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Question 11: 2 + 5 + 8 + 11 + \ldots + (3n-1) = \frac{1}{2} n(3n+1)

Answer:

Let P(n) : 2 + 5 + 8 + 11 + \ldots + (3n-1) = \frac{1}{2} n(3n+1)

For n = 1 , L.H.S = 2 R.H.S = 2

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 2 + 5 + 8 + 11 + \ldots + (3k-1) = \frac{1}{2} k(3k+1)

… … … … … i)

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

2 + 5 + 8 + 11 + \ldots + (3k-1) + [3(k+1)-1]

= \frac{1}{2} k (3k+1) + ( 3k + 2)

= \frac{3k^2+7k+4}{2}

= \frac{(3k+4)(k+1)}{2}

= \frac{1}{2} (k+1) [3(k+1) +1]

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

Hence by the principle of mathematical induction

2 + 5 + 8 + 11 + \ldots + (3n-1) = \frac{1}{2} n(3n+1) is true for all n \in N

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Question 12: 1.3 + 2.4 + 3.5 + \ldots + n(n+2) = \frac{1}{6} n(n+1)(2n+7)

Answer:

Let P(n) : 1.3 + 2.4 + 3.5 + \ldots + n(n+2) = \frac{1}{6} n(n+1)(2n+7)

For n = 1 , L.H.S = 3 R.H.S = 3

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1.3 + 2.4 + 3.5 + \ldots + k(k+2) = \frac{1}{6} k(k+1)(2k+7) … … … … … i)

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

1.3 + 2.4 + 3.5 + \ldots + k(k+2) + ( k+1)(k+3)

= \frac{1}{6} k(k+1)(2k+7) + ( k+1)(k+3)

= (k+1) \Big[ \frac{k}{6} (2k+7)+(k+3) \Big]

= (k+1) \Big[ \frac{2k^2+13k+18}{6} \Big]

= \frac{1}{6} (k+1) (2k+9)(k+2)

= \frac{1}{6} (k+1) [( k+1)+1] [2(k+1)+7]

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

Hence by the principle of mathematical induction

1.3 + 2.4 + 3.5 + \ldots + n(n+2) = \frac{1}{6} n(n+1)(2n+7) is true for all n \in N

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Question 13: 1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3}

Answer:

Let P(n) : 1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3}

For n = 1 , L.H.S = 3 R.H.S = 3

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1.3 + 3.5 + 5.7 + \ldots + (2k-1)(2k+1) = \frac{k(4k^2+6k-1)}{3}

 … … … … … i)

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

1.3 + 3.5 + 5.7 + \ldots + (2k-1)(2k+1) + [2(k+1)-1][2(k+1)+1]

= \frac{k(4k^2+6k-1)}{3} + (2k+1)(2k+3)

= \frac{4k^3+6k^2-k+ 12k^2+24k+9}{3}

= \frac{4k^3+18k^2+23k+9}{3}

= \frac{1}{3} (k+1) [4k^2+14k+9]

= \frac{1}{3} (k+1) [ 4(k+1)^2 + 6(k+1) - 1]

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

Hence by the principle of mathematical induction

1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) = \frac{n(4n^2+6n-1)}{3} is true for all n \in N

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Question 14: 1.2 + 2.3 + 3.4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3}

Answer:

Let P(n) : 1.2 + 2.3 + 3.4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3}

For n = 1 , L.H.S = 2 R.H.S = 2

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1.2 + 2.3 + 3.4 + \ldots + k(k+1) = \frac{k(k+1)(k+2)}{3}

 … … … … … i)

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

1.2 + 2.3 + 3.4 + \ldots + k(k+1) + (k+1)(k+2)

= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)

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

= \frac{1}{3} (k+1) (k+2) (k+3)

= \frac{1}{3} (k+1) (k+1+1) (k+1+2)

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

Hence by the principle of mathematical induction

1.2 + 2.3 + 3.4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3} is true for all n \in N

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Question 15: \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n}

Answer:

Let P(n) : \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n}

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

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^k} = 1 - \frac{1}{2^k} … … … … … i)

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

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^k} + \frac{1}{2^{k+1}}

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

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

= 1 - \frac{1}{2^{k+1}} \Rightarrow P(n) is true for n = k+1

Hence by the principle of mathematical induction

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n} is true for all n \in N

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Question 16: 1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 = \frac{1}{3} n(4n^2-1)

Answer:

Let P(n) : 1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 = \frac{1}{3} n(4n^2-1)

For n = 1 , L.H.S = 1 R.H.S = 1

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

\Rightarrow 1^2 + 3^2 + 5^2 + \ldots +(2 k-1)^2 = \frac{1}{3} k(4k^2-1)

… … … … … i)

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

1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 + [ 2(k+1)-1]^2

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

= \frac{1}{3} k (2k - 1)(2k+1) + (2k+1)^2

= (2k+1) \Big[ \frac{k}{3} ( 2k -1) + (2k+1) \Big]

= (2k+1) \big[ \frac{2k^2 - k + 6k + 3}{3} \Big]

= (2k+1) \big[ \frac{2k^2 + 5k + 3}{3} \Big]

= \frac{1}{3} (2k+1)(2k+3)(k+1)

= \frac{k+1}{3} [ 4k^2 + 8k + 3]

= \frac{k+1}{3} [ 4(k+1)^2 - 1]

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

Hence by the principle of mathematical induction

1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 = \frac{1}{3} n(4n^2-1) is true for all n \in N

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Question 17: a+ar+ar^2+ \ldots + ar^{n-1} = a \Big( \frac{r^n - 1}{r-1} \Big) , r \neq 1

Answer:

Let P(n) : a+ar+ar^2+ \ldots + ar^{n-1} = a \Big( \frac{r^n - 1}{r-1} \Big) , r \neq 1

For n = 1 , L.H.S = a R.H.S = a \Big( \frac{r^1 - 1}{r-1} \Big) = a

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

a+ar+ar^2+ \ldots + ar^{k-1} = a \Big( \frac{r^k - 1}{r-1} \Big) , r \neq 1 … … … … … i)

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

a+ar+ar^2+ \ldots + ar^{n-1} + ar^k

= a \Big( \frac{r^k-1}{r-1} \Big) + ar^k

= \frac{ar^k-a+ar^{k+1} - ar^k}{r-1}

= \frac{ar^{k+1} - 1}{r-1}

= \frac{a(r^{k+1} - 1)}{r-1}

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

Hence by the principle of mathematical induction

a+ar+ar^2+ \ldots + ar^{n-1} = a \Big( \frac{r^n - 1}{r-1} \Big) , r \neq 1 is true for all n \in N

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Question 18: a+ (a+d) + ( a+ 2d) + \ldots + (a+(n-1)d) = \frac{n}{2} [2a+ (n-1)d]

Answer:

Let P(n) : a+ (a+d) + ( a+ 2d) + \ldots + (a+(n-1)d) = \frac{n}{2} [2a+ (n-1)d]

For n = 1 , L.H.S = a R.H.S = \frac{1}{2} [2a+(1-1)d] = a

\therefore L.H.S = R.H.S Hence P(1) is true for n = 1

Let P(n) is true for n = k

a+ (a+d) + ( a+ 2d) + \ldots + (a+(k-1)d) = \frac{k}{2} [2a+ (k-1)d] … … … … … i)

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

a+ (a+d) + ( a+ 2d) + \ldots + (a+(k-1)d) + [ a + kd]

= \frac{k}{2} [ 2a + (k-1) d] + [a+kd]

= ak + \frac{k(k-1)d}{2} + a + kd

= a(k+1) + \frac{k^2d-kd+2kd}{2}

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

= \frac{1}{2} (k+1) [ 2a+kd]

= \frac{1}{2} (k+1) [ 2a + (k+1-1)d ]

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

Hence by the principle of mathematical induction

a+ (a+d) + ( a+ 2d) + \ldots + (a+(n-1)d) = \frac{n}{2} [2a+ (n-1)d] is true for all n \in N

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Question 19: 5^{2n} -1 is divisible by 24 for all n \in N

Answer:

Let P(n) : 5^{2n} -1 is divisible by 24 for all n \in N

For n = 1 , L.H.S = 25-1 = 24 which is divisible by 24

Hence P(1) is true for n = 1

Let P(n) is true for n = k

5^{2k} -1 is divisible by 24 for all k \in N

\therefore 5^{2k} -1 = 24 \lambda \Rightarrow 5^{2k} = 24 \lambda +1 … … … … … i)

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

\Rightarrow 5^{2(k+1)} - 1 = 25. 5^{2k} - 1 = 25 (24 \lambda +1) - 1 = 25 \times 24 \lambda + 24 \Rightarrow 24 ( 25 \lambda + 1) 24 \mu where \mu = 25 \lambda + 1

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

Hence by the principle of mathematical induction

5^{2n} -1 is divisible by 24 for all n \in N

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Question 20: 3^{2n} +7 is divisible by 8 for all n \in N

Answer:

Let P(n) : 3^{2n} +7 is divisible by 8 for all n \in N

For n = 1 , L.H.S =3^2+7=16 which is divisible by 8

Hence P(1) is true for n = 1

Let P(n) is true for n = k

3^{2k} +7 is divisible by 8 for all k \in N

\therefore 3^{2k}+7 = 8 \lambda \Rightarrow 3^{2k} = 8 \lambda -7 … … … … … i)

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

\Rightarrow 3^{2(k+1)} +7 = 9. 3^{2k} +7 = 9 (8 \lambda -7) +7 = 72 \lambda -56 \Rightarrow 8 ( 9 \lambda - 7) = 8 \mu where \mu = 9 \lambda - 7

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

Hence by the principle of mathematical induction

3^{2n} +7 is divisible by 8 for all n \in N

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Question 21: 5^{2n+2} -24n - 25 is divisible by 576 for all n \in N

Answer:

Let P(n) : 5^{2n+2} -24n - 25 is divisible by 576 for all n \in N

For n = 1 , L.H.S = 5^4 -24 - 25 = 625- 49 = 576 which is divisible by 576

Hence P(1) is true for n = 1

Let P(n) is true for n = k

5^{2k+2} -24k - 25 is divisible by 576 for all k \in N

\therefore 5^{2k+2} -24k - 25 = 576 \lambda … … … … … i)

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

\Rightarrow 5^{2(k+1)+2} -24(k+1) - 25

= 5^{2k+2} . 5^2 - 24k - 24 -25

= 25 . 5^{2k+2} - 24k-49

= 25 (576 \lambda + 24k + 25) - 24k - 49

= 25.576 \lambda + 600k + 625 - 24 k - 49

= 25. 576 \lambda + 576 k + 576

= 576 ( 25 \lambda + k + 1)

= 576 \mu where \mu = 25 \lambda + k + 1

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

Hence by the principle of mathematical induction

5^{2n+2} -24n - 25 is divisible by 576 for all n \in N

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Question 22: 3^{2n+2} -8n - 9 is divisible by 8 for all n \in N

Answer:

Let P(n) : 3^{2n+2} -8n - 9 is divisible by 8 for all n \in N

For n = 1 , L.H.S = 3^4 -8 - 9 = 81- 17 = 64 which is divisible by 8

Hence P(1) is true for n = 1

Let P(n) is true for n = k

3^{2k+2} -8k - 9 is divisible by 8 for all n \in N

\therefore 3^{2k+2} -8k - 9 = 8 \lambda \Rightarrow 3^{2k+2} = 8 \lambda +8k+9 … … … … … i)

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

\Rightarrow 3^{2(k+1)+2} -8(k+1) - 9

= 3^{2k+2} . 3^2 - 8k - 8 -9

= 9 . 3^{2k+2} - 8k-17

= 9 (8 \lambda + 8k + 9) - 8k - 17

= 72 \lambda + 72k + 81 - 8 k - 17

= 72 \lambda + 64 k + 64

= 8 ( 9 \lambda + 8k + 8)

= 8 \mu where \mu = 9 \lambda + 8k + 8

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

Hence by the principle of mathematical induction

3^{2n+2} -8n - 9 is divisible by 8 for all n \in N

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Question 23: (ab)^n = a^n b^n for all n \in N

Answer:

Let P(n) : (ab)^n = a^n b^n for all n \in N

For n = 1 , L.H.S = ab    RHS = ab Therefore LHS = RHS

Hence P(1) is true for n = 1

Let P(n) is true for n = k

(ab)^k = a^k b^k for all k \in N

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

\Rightarrow (ab)^{k+1} = (ab)^k (ab) = (a^k b^k) ( ab) = a^{k+1}b^{k+1} = (ab)^{k+1}

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

Hence by the principle of mathematical induction

(ab)^n = a^n b^n for all n \in N

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Question 24: n(n+1)(n+5) is a multiple of 3 for all n \in N

Answer:

Let P(n) : n(n+1)(n+5) is a multiple of 3 for all n \in N

For n = 1 , L.H.S = 1 \times 2 \times 6 = 12 which is divisible by 3

Hence P(1) is true for n = 1

Let P(n) is true for n = k

k(k+1)(k+5) is divisible by 3 for all k \in N

\therefore k(k+1)(k+5) = 3 \lambda \Rightarrow k^3+6k^2+5k = 3 \lambda … … … … … i)

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

\Rightarrow (k+1)(k+2)(k+6)

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

= k^3+6k^2+3k^2+6k+2k+12

= (k^3+6k^2+5k)+(3k^2+3k+12)

= 3 \lambda+3(k^2+k+4)

= 3 \mu where \mu = \lambda+(k^2+k+4)

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

Hence by the principle of mathematical induction

n(n+1)(n+5) is a multiple of 3 for all n \in N

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Question 25: 7^{2n}+2^{3n-3} . 3^{n-1} is divisible by 25 for all n \in N

Answer:

Let P(n) : 7^{2n}+2^{3n-3} . 3^{n-1} is divisible by 25 for all n \in N

For n = 1 , L.H.S = 49 + 2^0 \times 3^0 = 50 which is divisible by 25

Hence P(1) is true for n = 1

Let P(n) is true for n = k

7^{2k}+2^{3k-3} . 3^{k-1} is divisible by 25 for all k \in N

\therefore 7^{2k}+2^{3k-3} . 3^{k-1} = 25 \lambda … … … … … i)

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

\Rightarrow 7^{2(k+1)}+2^{3(k+1)-3} . 3^{(+1)k-1}

= 49 \times 7^{2k} + 2^{3k} 3^k

= 49 ( 25 \lambda - 2^{3k-3} . 3^{k-1} ) + 2^{3k} 3^k

= 49 \times 25 \lambda - 49 \Big( \frac{2^{2k}}{8} \Big) \Big( \frac{3^k}{3} \Big) + 2^{3k} 3^k

= 49 \times 25 \lambda -2^{3k} 3^k ( \frac{49}{24} -1)

= 49 \times 25 \lambda -2^{3k} 3^k \Big( \frac{25}{24} \Big)

= 25 \mu where \mu = \Big( 49 \lambda - \frac{2^{3k} 3^k}{24}  \Big)

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

Hence by the principle of mathematical induction

7^{2n}+2^{3n-3} . 3^{n-1} is divisible by 25 for all n \in N