Consider :\(P(k) : \sum _{ t=1 }^{ k-1 }{ t(t+1) } =\frac { k(k-1)(k+1) }{ 3 } ​​​​k\ge 2\) 
P(k):1.2+2.3+3.4+....+(k-1)k \(=\frac { k(k-1)(k+1) }{ 3 } \)
Now P(k+1):1.2+2.3+3.4+...+(k1)k+k(k+1)
\(=\frac { k(k-1)(k+1) }{ 3 } +k(k+1)\)
\(=\frac { k(k+1)(k+2) }{ 3 } k\ge 2\)

\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +....+\frac { 1 }{ (3k-1)(3k+2) } =\frac { k }{ 6k+4 } \)
Now P(k+1):\(\frac { 1 }{ 2.5 } +\frac { 1 }{ 5.8 } +\frac { 1 }{ 8.11 } +....+\frac { 1 }{ (3k-1)(3k+2) } +\frac { 1 }{ [3(k+1)-1][3(k+1)+2] } \)
=\(\frac { k }{ 2(3k+2) } +\frac { 1 }{ (3k+2)(3k+5) } \)
=\(\frac { { 3k }^{ 2 }+5k+2 }{ 2(3k+2)(3k+5) } +\frac { (3k+2)(k+1) }{ 2(3k+2)(3k+5) } =\frac { k+1 }{ 2(3k+5) } \)
=\(\frac { k+1 }{ 6(k+1)+4 } \)

Consider 
P(k):\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ k } \right) =k+1\)
P(k+1):\(\left[ \left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) .....\left( 1+\frac { 1 }{ k } \right) \right] \left( 1+\frac { 1 }{ k+1 } \right) \)
\(=(k+1)\left(\frac{k+2}{k+1}\right) \)
=(k+1)+1

Consider P(k) : 1+2+2²+.... +2^k = 2^k+1-1
Now P(+1):1+2+2² +..+2^k = 2^k+1 
=2^k+1-1+2^k+1
=2^(k+1)+1 -1

Consider P(k):2+2.3+2.3² + ....+2.3^k-1 = (3^k-1)
Now,P(k+1):2+2.3+2.3² + ....+2.3^k-1 +2.3^k
=(3^K -1)+2.3^k =3.3^k -1 =3^k+1-1

Step I Let P(n) be the given statement.
i.e P(n);\(2n+1 <{ 2 }^{ n }\)
Step II For n = 3,we have
(2 x 3 +1)<2³ \(\Rightarrow \) 7< 8, which is true.
Thus P(1) is true.
Step III Let us assume that P(k) is true.
i.e P(k):\(2k+1 <{ 2 }^{ k }\)
Step IV Now, we shall prove the statement for n=k+1.
for this, we have to show that \(2(k+1)+1<{ 2 }^{ k+1 }\)
from Eq.(i), \(2k+1 <{ 2 }^{ k }\)
So, (2k+1)+2 < 2^k +2 [adding 2 on both sides]
\(\Rightarrow 2 k+3<2^{k} \cdot 2 \quad {\left[\because 2^{k}+2<2^{k} \cdot 2\right]} \)
\(\Rightarrow 2 k+3<2^{k+1} \Rightarrow 2(k+1)+1<2^{k+1}\)
thus, P(k+1) is true, whenever P(k) is true.
hence, by principle of mathematical induction P(n) is true for all natural numbers, \(n\ge 3\).

Step I Let P(n) be the given statement, i.e.
\(P(n):\quad 1+2+3+.....+n=\frac { n(n+1) }{ 2 } \)
Step II For n=1, we have; LHS=1 and
RHS=\(\frac { 1.(1+1) }{ 2 } =\frac { 1\times 2 }{ 2 } =1\)
 \(\therefore \) LHS=RHS   P(1) is true
Step III Let us assume that P(n) is true for n=k. Then we have
P(k): \(1+2+3+.....+K=\frac { k(k+1) }{ 2 } \quad \quad \quad ....(1)\)
Step IV Now, we shall prove the statement for n=k+1
For this we have to show that
\(1+2+...+(k+1)=\frac { (k+1)(k+1+1) }{ 2 } \)
Consider, LHS=1+2+.....+k+(k+1)
\(=\frac { k(k+1) }{ 2 } +(k+1)\)   [using Eq.(i)]
\(=(k+1)\left( \frac { k }{ 2 } +1 \right) \)    [taking common (k+1)]
\(=\frac { (k+1)(k+2) }{ 2 } =\frac { (k+1)(k+1+1) }{ 2 } =RHS\)
Thus, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all natural numbers n.

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

Consider \(P(k) : \frac { { k }^{ 7 } }{ 7 } +\frac { { k }^{ 5 } }{ 5 } +\frac { 2 }{ 3 } k^{ 3 }-\frac { k }{ 105 } \lambda \in I\)
Now,
P(k+1):\(\frac { ({ k+1) }^{ 7 } }{ 7 } +\frac { { (k }+1)^{ 5 } }{ 5 } +\frac { 2 }{ 3 } (k+1)^{ 3 }-\frac { k+1 }{ 105 } \)
\(=\frac { 1 }{ 7 } ({ k }^{ 7 }+{ 7k }^{ 6 }+{ 21k }^{ 5 }{ +35k }^{ 4 }{ +35k }^{ 3 }+{ 21k }^{ 2 }{ +7k }+1)+\frac { 1 }{ 5 } ({ k }^{ 5 }+{ 5k }^{ 4 }+{ 10k }^{ 3 }{ +10k }^{ 2 }{ +5k })+{ 1 }\)
\(+\frac { 2 }{ 3 } ({ k }^{ 3 }+{ 3k }^{ 2 }+{ 3k }+{ 1 })-\frac { k+1 }{ 105 } \)
=\(+\lambda +({ k }^{ 6 }+{ 3k }^{ 5 }+{ { 6k }^{ 4 }+{ 7k }^{ 3 } }+{ { 7k }^{ 2 }+4k })\)= Integer

Consider  \(P(k) : { 4 }^{ k }=3\lambda +1\)
Now,
P(k+1) :\({ 4 }^{ k+1 }={ 4 }^{ k }.4=(3\lambda +1)4\)
\(=12\lambda +4=3(4\lambda +1)+1\)

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.

Step I: Let P(n) be the given statement
i.e. P(n): \(1\times 1!+2\times 2!+3\times 3!+....+n\times n!=(n+1)!-1\)
Step II : For n=1, we have
LHS=\(1\times 1!\)=1
and RHS=(1+1)!-1=2!-1=2-1=LHS
\(\because \)  LHS=RHS
\(\therefore \)  P(1) is true
Step III  Let us assume that P(n) is true for n=k
Then, we have
P(k): \(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)!-1\quad \quad ....(i)\)
Step IV  Now, we shall prove the statement for n=k+1. For this we have to show that
\(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!=(k+1+1)!-1\)
Then, LHS \(=1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!\)
=(k+1)!-1+(k+1)!\(\times \)(k+1)      [from Eq.(1)]
=(k+1+1)(k+1)!-1=(k+2)(k+1)!-1
=(k+2)!-1   [\(\because \)n(n-1)!=n]
Thus, P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n.

Consider P(k) :\(1.4.7+2.5.8+3.6.9+....+k(k+3)(k+6)=\frac { k }{ 4 } (k+1)(k+6)(k+7)\)
Now, P(k+1)1.4.7+2.5.8+3.6.9+.....+k(k+3)(k+6)+(k+1(k+4)(k+7)
=\(\frac { k }{ 4 } \)(k+1)(k+6)(k+7)+(k+1)(k+4)(k+7)
=\(\frac { (k+1)(k+7) }{ 4 } ({ k }^{ 2 }+10k+16)\)
=(k+1)(k+7)(k+8)(k+2)

Step I  Let P(n) be the given statement.
i.e. P(n): \(1+3+3^{ 2 }+.....+3^{ n-1 }=\frac { 3^{ n }-1 }{ 2 } \)
Step II  For n=1, we have
LHS=3^1-1=3⁰=1 and RHS= \(\frac { 3^{ 1 }-1 }{ 2 } =1\)
\(\therefore \)   LHS=RHS 
\(\therefore \)   P(1) is true.
Step III  Let us assume that P(n) is true for n=k
Then, we have
\(P(k) :1+3+3^{ 2 }+.....+3^{ k-1 }=\frac { 3^{ k }-1 }{ 2 } ....(i)\)    
Step IV  Now, we shall prove the statement for n=k+1 For this,                  
we have to show
\(1+3+3^{ 2 }+.....+3^{ k-1 }+3^{ k }=\frac { 3^{ k+1 }-1 }{ 2 } \)
Then, LHS=1+3+3²+...3^k-1+3^k
\(=\frac { 3^{ k }-1 }{ 2 } +{ 3 }^{ k }\)      [from Eq. (i)]
\(=\frac { 3^{ k }-1+2.3^{ k } }{ 2 } =\frac { 3^{ k }(1+2)-1 }{ 2 } \)
\(=\frac { 3^{ k }.3-1 }{ 2 } =\frac { 3^{ k+1 }-1 }{ 2 } =RHS\)
Thus, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n \in N\)

Step I Let P(n) be the given statement.
i.e.P(n) :\({ (1+x) }^{ n }\ge (1+nx)\)
Step II for n=1, we have \({ (1+x) }\ge (1+x)\), which is true.Thus P(1) is true.
Step III Let us assume that P(k) is true.
i.e P(k) : \({ (1+x) }^{ k }\ge (1+kx)\)  ....(i)
Step IV Now, we shall prove the statement for n=k+1.For this, we have to show that
\({ (1+x) }^{ k+1 }\ge (1+(k+1)x)\)
from Wq (i) we have \({ (1+x) }^{ k }\ge (1+kx)\)....(ii)
\(\because \) x > 1  \(\Rightarrow \)   x+1>0
So, on multiplying both sides of Eq.(ii) by (x+1), we get
\({ (1+x) }^{ k }(1+x)\ge (1+kx)(1+x)\)
(1+x)^{k+1 \(\ge \)} 1+x+kx+kx² .....(iii)
Here, k is a natural number and \({ x }^{ 2 }\ge 0\), therefore  \(k{ x }^{ 2 }\ge 0\) and so, \((1+x+kx+k{ x }^{ 2 })\ge (1+x+kx)\)
Then, from Eq(iii), we have 
\({ (1+x) }^{ k+1 }\ge (1+x+kx)\)
or \({ (1+x) }^{ k+1 }\ge [1+(1+k)x]\)
Thus,P(k+1) is true whenever P(k) is true.Hence, by the principle mathematical induction, P(n) is true for all natural numbers.