\({ 1 }^{ 2 }{ +2 }^{ 2 }+....+{ k }^{ 2 }>\frac { { k }^{ 3 } }{ 3 } \)
\(\Rightarrow \) \( ​​​​{ 1 }^{ 2 }{ +2 }^{ 2 }+....+{ k }^{ 2 }+{ (k+1) }^{ 2 }>\frac { { k }^{ 3 } }{ 3 } +{ (k+1) }^{ 2 }\)
\(=\frac { 1 }{ 3 } [{ k }^{ 3 }+{ S(k+1) }^{ 2 }]=\frac { 1 }{ 3 } [{ k }^{ 3 }{ +3k }^{ 2 }+6k+3]\)
\(=\frac { 1 }{ 3 } [({ k }+1)^{ 3 }+3k+2]\)
\({ 1 }^{ 2 }{ +2 }^{ 2 }+....+{ k }^{ 2 }>\frac { 1 }{ 3 } (k+1){k}^{3}\)

2k<(k+2)!\(\Rightarrow \)2k+2(k+2)!+2]
\(\Rightarrow \) (k+1)² < 2^k + 2^k    [(2k+1)<2^k for k\(\le \)3]
\(\Rightarrow \) (k+1)² < 2^k+1

Consider P(k): (2k+7)<(k+3)²
\(\Rightarrow \) (2k+7)+2 < (k+3)² +2
\(\Rightarrow \) 2(k+1)+7 < (k+4)²  [(k+3)³+2<(k+4)²]

Let P(n) = (xⁿ-yⁿ) is divisible by (x - y) for all n ∊ N.
For n =1
P(1) = (x¹-y¹) is divisible by (x-y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k)=(x^k-y^k) is divisible by (x-y)
⇒ (x^k-y^k) =m(x-y) for some m ∊ Z     ..(i)
For n = k+1
∴  P(k+1)=x^k+1-y^k+1 is divisible by (x-y)
x^k+1-y^k+1 = x^k+1-x^ky+x^ky-y^k+1
= x^k(x-y)+y(x^k-y^k)
= x^k(x -y)+y.m(x-y)
= (x-y)[x^k+my]
which is divisible by (x-y)
∴  P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, pen) is true for all n∊N.

Consider P(k):\({ k }^{ 3 }-7k+3=3\lambda \)
Now,P(k+1):\({ (k+1) }^{ 3 }-7(k+1)+3\)
= k³+3k²+3k+1-7k-4
= \((3\lambda -3)+{ 3k }^{ 2 }+3k-3=3({ k }^{ 2 }+k-2)\)

Let P(n) = sin x +sin 2x+sin 3x+..+sin nx = \(\frac { sin\left( \frac { n+1 }{ 2 } \right) x\quad sin\frac { nx }{ 2 } }{ sin\frac { x }{ 2 } } \)
For n =1 P(1) = sin x =     ⇒ sin x =sin x
∴ P(1) is true
Let p(n) be true for n = k
∴  P(k) =sin k+sin 2k+sin 3k+..+sin nk=    \(\frac { sin\left( \frac { k+1 }{ 2 } \right) x\quad sin\frac { kx }{ 2 } }{ sin\frac { x }{ 2 } } \)...(i)
For n = k+1
∴  P(k + 1) = sinx + sin 2x+ sin 3x+ ...+ sin kx + sin (k + 1) x= \(\frac { sin\left( \frac { k+2 }{ 2 } \right) x\quad sin\frac { (k+1) }{ 2 } }{ sin\frac { x }{ 2 } } \)
= \(\frac { sin\left( \frac { k+1 }{ 2 } \right) x\quad sin\frac { kx }{ 2 } }{ sin\frac { x }{ 2 } } +sin\quad (k+1)x\)
= \(\frac { sin\left( \frac { k+1 }{ 2 } \right) x\quad sin\frac { kx }{ 2 } }{ sin\frac { x }{ 2 } } +2\quad sin\left( \frac { k+1 }{ 2 } \right) x\quad cos\left( \frac { k+1 }{ 2 } \right) x\)
= \(sin\left( \frac { k+1 }{ 2 } \right) x\left[ \frac { sin\frac { kx }{ 2 } }{ sin\frac { x }{ 2 } } +2cos\left( \frac { k+1 }{ 2 } \right) x \right] \)
= \(sin\left( \frac { k+1 }{ 2 } \right) x\left[ \frac { sin\frac { kx }{ 2 } +2\quad sin\frac { x }{ 2 } cos\left( \frac { k+1 }{ 2 } \right) x }{ sin\frac { x }{ 2 } } \right] \)
=  \(sin\left( \frac { k+1 }{ 2 } \right) x\left[ \frac { sin\frac { kx }{ 2 } +2sin\left( \frac { k+2 }{ 2 } \right) x-sin\frac { kx }{ 2 } }{ sin\frac { x }{ 2 } } \right] \)               [∵ 2sinA cos A=sin(A+B)-sin(A-B)]
= \(\left[ \frac { sin\left( \frac { k+1 }{ 2 } \right) x.sin\left( \frac { k+2 }{ 2 } \right) x }{ sin\frac { x }{ 2 } } \right] \)
Which is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

P(n) = 3ⁿ< 2ⁿ
For n =1 ,P(1)= 31 > 21 ⇒ 3 > 2
∴ P(1) is true
Let P(n) is true for n = k
∴  P(k) =3^k<2^k
For n =k + 1
P(k+1) = 3^k+1>2^k+1
From (i), 3^k> 2^k
Multiplying by 3 on both sides
3^k.3 >2^k.3 ⇒ 3^k+1>2^k(2+1)  ⇒ 3^k+1>2^k+1+2^k ⇒ 3^k+1>2^k+1 because k²>0
∴  P(k+1) is true

Let P(n) = 4ⁿ-14ⁿ is a multiple of 27
For n =1
P(1) = 41¹-14¹ is a multiple of 27
⇒ 27 is a multiple of 27
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = 41^k-14^k is a multiple of 27
⇒ 41^k-14^k= 17λ     ..(i)
For n = k+ 1
P(k + 1) = 41^k+1-14^k+1 is a multiple of 27
Now 41^k+1-14^k+1
= 41^k+1-41^k.14+41^k.14-14^k+1
= 41^k(41-14)+14(41^k-14^k)
= 41^k\(\times\)27+14\(\times\)27λ      [Using (i)]
= 27(41^k+14λ)
⇒ 41^k+1-14^k+1 is a multiple of 27
∴ P(k+1) is true
Thus P(k) is true ⇒P(k+1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

Let P(n) = 3²ⁿ⁺¹-8n-9is divisible by 8
For n =1
P(1) = 3^{2\(\times\)1+2}-81-9 is dividible by 8
⇒ 3⁴-8-9 is divisible by 8
⇒ 64 is divisible by 8
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = 3^2k+2- 8k-9 is divisible by 8
⇒ 3^2k+2-8k-9 = 8λ
⇒  3^2k+2= 8λ+8k+9 ...  (i)
For n = k+1
P(k+1) = 3^2(k+1)+2- 8(k+1)-9 is divisible by 8
= 3^2(k+1).3²-8k-8-9
= (8λ+8k+9)9-8k-17
= 72λ+72k+81-8k-17   [Using (i)]
= 72λ+64k+64=8(9λ+8k+8)
⇒  3^2(k+1)+2-8(k+1)-9 is divisible by 8
∴ P(k + 1) is true
Thus P(k) is true  ⇒ P(k+1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

Let P(n)
\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ n } \right) \)=(n+1)
For n =1
P(1) =\(\left( 1+\frac { 1 }{ 1 } \right) \)=  1+1⇒2=2
∴ P(1) is true
Let P(n) be true for n = k
∴ p(k) = \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{k } \right) \)   ...(i)
For n = k+1
R.H.S =(k+2)
L.H.S = (k+1)  \(\left( 1+\frac { 1 }{ k+1 } \right) \) [Using (i)]
= (k+1) \(\left[ \frac { k+1+1 }{ k+1 } \right] \) =(k+2)
∴ P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊ N.

Step I : Let P(n) be the given statement.
i.e., P(n): \(\left( 1+\frac { 3 }{ 1 } \right) \left( 1+\frac { 5 }{ 4 } \right) \left( 1+\frac { 7 }{ 9 } \right) ...\left( 1+\frac { (2n+1) }{ { n }^{ 2 } } \right) =(n+1)^{ 2 }\)
Step II : For n =1
\({ LHS }=\left(1+\frac{3}{1}\right)=4 \)
\({ and }\quad \mathrm{RHS}=(1+1)^{2}=2^{2}=4 \)
\(\because \quad \mathrm{LHS}=\mathrm{RHS}\)
∴ P(1) is true
Step III : Let as assume that pen) is true for n = k.Then, we have
= \(\left( 1+\frac { 3 }{ 1 } \right) \left( 1+\frac { 5 }{ 4 } \right) \left( 1+\frac { 7 }{ 9 } \right) ...\left( 1+\frac { (2k+1) }{ { k }^{ 2 } } \right) \)   ...(i)
= (k+1)²
Step IV : Now, we shall prove the statement for n = k + 1.
For this, we have to show that
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 k+1}{k^{2}}\right)\)
\({\left[1+\frac{2(k+1)+1}{(k+1)^{2}}\right]=(k+1+1)^{2}}\)
Then, \(\text { LHS }=\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 k+1}{k^{2}}\right) {\left[1+\frac{2(k+1)+1}{(k+1)^{2}}\right]}\)
\(=(k+1)^{2}\left[1+\frac{2(k+1)+1}{(k+1)^{2}}\right] \quad[\text { from } \mathrm{Eq}]\)
\(=(k+1)^{2}\left[\frac{(k+1)^{2}+2(k+1)+1}{(k+1)^{2}}\right]\)
\(=(k+1)^{2}\left[\frac{(k+1)^{2}+2 k+3}{(k+1)^{2}}\right]\)
\(=(k+1)^{2}+2 k+3=k^{2}+2 k+1+2 k+3\)
\(=k^{2}+4 k+4=(k+2)^{2}=(k+1+1)^{2}=\mathrm{RHS}\)
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.