Question 1: Let $P (n)$ be the statement $7$ divides $(2^{3n} -1)$. What is $P (n + 1)$ ?

$P (n + 1)$ is the statement $7$ divides $(2^{3(n+ 1)} - 1)$.

Clearly, $P (n + 1)$ is obtained by replacing $n$ by $(n+ 1)$ in $P (n)$.

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Question 2: If $P (n)$ is the statement $n^2 > 100$, prove that whenever $P(r)$ is true, $P (r +1)$ is also true.

The statement $P (n)$ is $n^2 >100$. Let $P (r)$ be true. Then $r^2 >100$

We wish to prove that the statement $P (r +1)$ is true i.e. $(r +1)^2 >100$

Now,

$P (r)$ is true

$\Rightarrow r^2 > 100$

$\Rightarrow r^2 + 2r + 1> 100 + 2r + 1$ [ adding $(2r+1)$ on both sides]

$\Rightarrow (r+1)^2 > 100 + 2r + 1$

$\Rightarrow (r+1)^2 > 100$

$\Rightarrow P(r+1)$ is true [ $\because 100 + 2r + 1 > 100$ for every natural number $r$ ]

Thus, whenever $P(r)$ is true, $P (r + 1)$ is also true.

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Question 3: Let $P(n)$ be the statement $3^n > n$. If $P(n)$ is true, prove that $P(n+1)$ is true.

We are given that $P (n)$ is true i.e. $3^n >n$ and we wish to prove that $P (n + 1 )$ is true i.e. $3^(n+ 1) >(n+1)$.

Now,

$P (n)$ is true

$\Rightarrow 3^n > n$

$\Rightarrow 3.3^n > 3n$ [Multiplying both sides by $3$]

$\Rightarrow 3^{n+1} > n + 2n$

$\Rightarrow 3^{n+1} > n + 1$ [ $\because 2n> 1$ for every $n \in N \Rightarrow 2n+n>n+1$. for every $n \in N$ ]

$\Rightarrow P (n+1)$ is true

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Question 4: lf $P (n)$ is the statement $2^{3n} - 1$ is an integral multiple of $7$, and if $P(r)$ is true, prove that $P (r + 1)$ is true.

Let $P (r)$ be true. Then, $2^{3r} -1$ is an integral multiple of $7$.

We wish to prove that $P (r + 1)$ is true i.e. $2^{3(r+ 1)} - 1$ is an integral multiple of $7$.

Now,

$P(r)$ is true

$\Rightarrow 2^{3r}-1$. is an integral multiple of $7$

$\Rightarrow 2^{3r}-1 = 7 \lambda$, for some $\lambda \in N$

$\Rightarrow 2^{3r} = 7 \lambda + 1$

Now, $2^{3(r + 1)}-1 = 2^{3r} \times 2^3 -1, = (7 \lambda +1) \times 8-1$

$\Rightarrow 2^{3(r+1)} -1 = 56 \lambda +8-1 = 56 \lambda +7 =7 (8\lambda+1)$

$\Rightarrow 2^{3(r+1)}-1 =7 \mu$, where $\mu =8 \lambda +1 \in N$

$\Rightarrow 2^{3(r + l)}-1$ is an integral multiple of $7$

$\Rightarrow P(r+1)$ is true

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Question 5: Prove by the principle of mathematical induction that for all $n \in N, n^2 + n$ is even natural number.

Let $P (n)$ be the statement $n^2 + n$ is even

We have, $P (n): n^2 + n$ is even

$\because 1^2 +1 = 2$, which is even

$\therefore P (1)$ is true

Let $P (m)$ be true. Then,

$P(m)$ is ture $\Rightarrow ,m^2 + m$ is even $\Rightarrow m^2 +m = 2\lambda$ for some $\lambda \in N$

Now, we shall show thal $P (m+ 1)$ is true. For this we have to show that $(m + 1)^2 + (m+1)$ is an even natural number.

$\Rightarrow (m+1)^2 +(m+1) = (m^2 +2m + 1) + (m+1) = (m^2 + m )+(2m+2)$

$\Rightarrow (m+1)^2 +(m+1) = m^2 + m + 2(m+1) = 2\lambda +2(m+1)$

$\Rightarrow (m+1)^2 +(m+1) = 2( \lambda +m+1) = 2\mu$ ,where $\mu = \lambda +m+1 \in N$

$\Rightarrow (m + 1)^2 + (m + 1)$ is an even natural number

$\Rightarrow P(m+1)$ is true

Thus, $P (m)$ is true $\Rightarrow P (m + 1)$ is true

Hence,by the principle of mathematical induction, $P (n)$ is true for all $n \in N$ i.e. $n^2 +n$ is even for all $n \in N$.

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Question 6: Prove by the principle of mathematical induction that $n (n + 1) (2n + 1)$ is divisible by $6$ for all $n \in N$

Let $P (n)$ be the statement $n (n + 1) (2n + 1)$ is divisible by $6$

i.e. $P (n) : n (n + 1) (2n + 1)$ is divisible by $6$

We have, $P (1) : 1 (1 + 1) (2 + 1)$ is divisible by $6$

$\because 1 (1 + 1) (2 + 1) = 6$ which is divisibleby $6$

$\therefore P(1$) is true

Let $P(m)$ be true. Then,

$m (m + 1) (2m + 1)$ is divisible by $6$

$m (m + 1) (2m + 1) = 6 \lambda$, for some $\lambda \in N$

Now, we shall show that $P (m + 1)$ is true. For this we have to show that $(m+1) (m + 1 + 1) [ 2(m + 1) + 1 ]$ is divisibleby $6$

$(m+1)(m+ 1 + 1) [ 2(m+1) + 1 ] = (m+1)(m+2) [(2m+ 1) + 2 ]$

$= (m + 1) (m + 2) (2m + 1) + 2 (m + 1) (m + 2)$

$= m (m + 1) (2m + 1) + 2 (m + 1) (2m + 1) + 2 (m + 1) (m + 2)$

$= m (m + 1) (2m + 1) + 2 (m + 1) (2m + 1 + m + 2)$

$= m (m + 1) (2m + 1) + 2 (m + 1) (3m + 3)$

$= m(m+1)(2m+1)+6 (m+1)^2 = 6\lambda + 6(m+1)^2$

$= 6 { \lambda + ( m+1)^2 }$ which is divisible by $5$

$\Rightarrow P (m + 1)$ is true

Thus, $P (m)$ is true $\Rightarrow P (m + 1)$ is true

Hence, by the principle of mathematical induction, the given statement is true for all $n \in N$

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Question 7: Prove by induction that the sum $S_n =n^3 + 3n^2 + 5n + 3$ is divisible by $3$ for all $n \in N$

Let $P (n)$ be the statement given by $P(n): S_n =n^3 + 3n^2 + 5n + 3$ is divisible by $3$

Wehave, $P(1):S_1 = 1^3 + 3(1)^2 +5(1) + 3 = 12$ is divisible by $3$

$\therefore P (1)$ is true

Let $P(m)$ be true. Then,

$S_n =m^3 + 3m^2 + 5m + 3$ is divisible by $3$

$S_n =m^3 + 3m^2 + 5m + 3 = 3 \lambda$ for $\lambda \in N$

We now wish to show that $P (m+ 1)$ is true. For this we have to show that $(m +1)^3 + 3 (m+ 1)^2 + 5 (m +1) + 3$ is divisible by $3$

Now, $(m +1)^3 + 3 (m + 1)^2 + 5 (m+ 1) + 3$

$= (m^3 + 3 m^2 + 5 m + 3) + 3m^2 + 9m + 9$

$= 3 \lambda + 3(m^2 + 3m+ 3)$

$= 3( \lambda + m^2+3m+3)$

$= 3 \mu$ where $\mu = \lambda + m^2+3m+3 \in N$

$\therefore P (m + 1)$ is true

Thus, $P (m)$ is true $\Rightarrow P (m + 1)$ is true

Hence, by the principle of mathematical induction the statement is true for all $n \in N$

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Question 8: Using principle of mathematical induction, prove that $x^{2n} - y^{2n}$ it divisible by $x + y$ for all $n \in N$

Let $P (n)$ be the statement given by $P(n) : x^{2n} - y^{2n}$ is divisible by $(x + y)$

$P (1) = (x^2 - y^2) = (x-y)(x+y)$ which is divisible by $(x + y)$

So, $P(1)$ is true.

Let $P (m)$ be true. Then,

$x^{2m} - y^{2m}$ is divisible by $x + y$

$\Rightarrow x^{2m} - y^{2m} = \lambda ( x+y)$

We shall now show that $P (m+ 1)$ is true i.e. $x^{2m+2} - y^{2m+2}$ is divisible by $(x + y)$

Now $x^{2m+2} - y^{2m+2} = x^{2m+2} - x^{2m}y^2 + x^{2m}y^2 - y^{2m+2}$

$= x^{2m} (x^2 - y^2) + y^2 ( x^{2m} - y^{2m})$

$= x^{2m} (x^2 - y^2) + y^2 \lambda (x+y)$

$= (x+y) [ x^{2m} (x-y) + \lambda y^2 ]$

Clearly, it is divisible by $(x + y)$

$\therefore P (m + 1)$ is true

Thus, $P (m)$ is true $\Rightarrow P (m + 1)$ is true.

Hence , by the principle of mathematical induction $P (n)$ is true for all $n \in N$

i.e. $x^{2n} - y^{2n}$ it divisible by $(x + y)$ for all $n \in N$

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Question 9: Using the principle of mathematical induction, prove that $(2^{3n}-1)$ is divisible by $7$ for all $n \in N$

Let $P (n)$ be the statement given by $P (n) : 2^{3n} -1$ is divisible by $7$

$P (1) :2^{3 \times 1} - 1$ is divisible by $7$

Clearly, $2^{3 \times 1} -1= 8 - 1 = 7$ which is divisibte by $7$

So, $P (1)$ is true

Let $P (m)$ be true. Then,

$2^{3m} -1$ is divisible by $7$

$\Rightarrow 2^{3m} -1 = 7 \lambda$ for $\lambda \in N$

We shall now show that $P (m+ 1)$ is true. For this we have to show that $2^{3(m + 1)} -1$ is divisible by $7$

Now, $2^{3(m+ 1)} -1 = 2^{3m} \times 2^3 -1 = (7 \lambda + 1) 2^3 -1 = 56 \lambda + 8 - 1 = 7 (8 \lambda + 1)$ is divisible by $7$

$\therefore P (m+ 1)$ is true

$\therefore P (m)$ is true $= P (m + 1)$ is true

Hence , by the principle of mathematical induction, $P(n)$ is true for all $n \in N \Rightarrow (2^{3n}-l)$ is divisible by $7$ for all $n \in N$

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Question 10: Prove that: $2.7^n + 3.5^n - 5$ is divisible by $24$, for all $n \in N$

Let $P (n)$ be the statement given by $P(n) : 2.7^n + 3.5^n - 5$ is divisible by $24$, for all $n \in N$

We have, $P(1) = 2.7^1 + 3.5^1 - 5 = 14 + 15 - 5 = 24$ which is divisible by $24$

$\therefore P (1)$ is true

Let $P (m)$ be true. Then, $2.7^m + 3.5^m - 5$ is divisible by $24$, for all $m \in N$

$\Rightarrow 2.7^m + 3.5^m - 5 = 24 \lambda$ for some $\lambda \in N$

$\Rightarrow 3.5^m = 24 \lambda + 5 - 2.7^m$

We shall now show that $P (m+ 1)$ is true. For this we have to show that $2.7^{m+1} + 3.5^{m+1} - 5$ is divisible by $24$

$2.7^{m+1} + 3.5^{m+1} - 5$

$= 2.7^{m+1} + (3.5^m)5 - 5$

$= 2.7^{m+1} + (24 \lambda + 5 - 2.7^m)5 - 5$

$= 2.7^{m+1} + 120 \lambda + 25 - 10. 7^m - 5$

$= ( 2.7^{m+1} - 10. 7^m) + 120 \lambda + 20$

$= ( 2.7^{m+1} - 10. 7^m) + 120 \lambda + 24-4$

$= ( 14 - 10) 7^m - 4 + 24 ( 5 \lambda + 1)$

$= 4 ( 7^m - 1) + 24 ( 5 \lambda + 1)$

$= 4 \times 6\mu + 24 ( 5 \lambda + 1)$

[ Since $7^m - 1$ is a multiple of $6$ for all $m \in N \therefore 7^m - 1 = 6 \mu$ for all $\mu \in N$ ]

$= 24 ( \mu + 5 \lambda + 1 )$ which is divisible by $24$

Therefore $P (m + 1)$ is true

Thus, $P (m)$ is true $\Rightarrow P (m + 1)$ is true

Hence, by the principle of mathematical induction, $P (n)$ is true for all $n \in N$

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Question 11: Prove by induction that $(2n+7)<(n+3)^2$ for all natural numbers $n$. Using this, prove by induction that $(n+ 3)^2 \leq 2^{n+3}$ for all $n \in N$

Let $P (n)$ be the statement given by $P(n) : (2n+7)<(n+3)^2$

$P(1) = ( 2 \times 1 + 7) < ( 1 + 3 )^2 \Rightarrow 9 < 16$

Therefore $P (1)$ is true

Let $P (m)$ be true. Then, $(2m+7)<(m+3)^2$ for all natural numbers $m$

We shall now show that $P (m + 1)$ is true whenever $P (m)$ is true. For this we have to show that $(2(m+1)+7)<((m+1)+3)^2$

Now, $P (m)$ is true

$(2m+7)<(m+3)^2$

$\Rightarrow (2m+7) + 2 <(m+3)^2 + 2$

$\Rightarrow 2(m+1) +7 < m^2 + 6m+11$

$\Rightarrow 2(m+1)+7

$\Rightarrow 2(m+1)+7 <(m+4)^2$

$\Rightarrow [2(m+1) +7] < [(m+1) + 3]^2$

$\Rightarrow P (m + 1)$ is true

Hence, by the principle of mathematical induction $P (n)$ is true for all $n \in N$

Let $P'(n)$ be the statement given by $P'(n) :(n+ 3)^2 \leq 2^{n+3}$ for all $n \in N$

$P'(1) : (1 + 3)^2 \leq 2^{1+ 3}$

$\because (1 + 3)^2 = 16 \leq 2^{1+ 3}$

$\therefore P'(1)$ is true

Let $P'(m)$ be true. Then, $(m+ 3)^2 \leq 2^{m+3}$ for all natural numbers $m$

We shall now show that $P' (m + 1)$ is true whenever $P' (m)$ is true. For this we have to show that $((m+1)+3)^2 \leq 2^{(m+1)+3}$

Now, $P' (m)$ is true

$\Rightarrow (m+ 3)^2 \leq 2^{m+3}$

$\Rightarrow (m+ 3)^2 + ( 2m + 7) \leq 2^{m+3} + ( 2m + 7)$

$\Rightarrow (m+4)^2 \leq 2^{m+3} + ( m+3)^2$

$\Rightarrow (m+4)^2 \leq 2^{m+3} + 2^{m+3}$

$\Rightarrow (m+4)^2 \leq 2.2^{m+3}$

$\Rightarrow (m+4)^2 \leq 2^{m+4}$

$\Rightarrow [ (m+1)+3]^2 \leq 2^{(m+1)+3}$

$\Rightarrow P'(m+1)$ is true

Hence by the principle of mathematical induction $P'(n)$ is true for all $n \in N$ i.e. $(n+ 3)^2 \leq 2^{n+3}$ for all $n \in N$