Let \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ (n+1) } \)
For n =1
 P(1) = \(\frac { 1 }{ 1(1+1) } =\frac { 1 }{ 1+1 } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ (k+1) } \)   ...(i)
For n = k+1
∴ P(k+1)= \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } +\frac { 1 }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)
= \(\frac { k }{ k+1 } +\frac { 1 }{ (k+1)(k+2) } =\frac { k(k+2)+1 }{ (k+1)(k+2) } =\frac { { k }^{ 2 }+2k+1 }{ (k+1)(k+2) } \)       [Using (i)]
=  \(\frac { (k+1)^{ 2 } }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)= 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.

Let P(n)= \(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +....+\frac { 1 }{ 2^{ n } } =1-\frac { 1 }{ 2^{ n } } \)
For n =1
P(1)=\(\frac { 1 }{ 2^{ 1 } } =1-\frac { 1 }{ 2^{ 1 } } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) =\(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +....+\frac { 1 }{ 2^{ k } } \)
=  \(1-\frac { 1 }{ 2^{ k } } \)  ........(i)
For n = k + 1
P(k+1) = \(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +....+\frac { 1 }{ 2^{ k } } +\frac { 1 }{ 2^{ k+1 } } \)
= \(1-\frac { 1 }{ 2^{ k+1 } } \)
=  \(1-\frac { 1 }{ 2^{ k } } +\frac { 1 }{ 2^{ k+1 } } \)    [Using (i)]
= 1- \(\left( \frac { 1 }{ 2^{ k } } -\frac { 1 }{ 2^{ k+1 } } \right) \)
=1-\(\left( \frac { 2-1 }{ 2^{ k+1 } } \right) =1-\frac { 1 }{ 2^{ k+1 } } \)
∴ 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) = 1.2+2.3+3.4+....+n(n+1) = \(\left[ \frac { n(n+1)(n+2) }{ 3 } \right] \)
For n =1
P(1) =1(1+1)
⇒ 2=2
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = 1.2 + 2.3 +3.4 +... + k(k + 1)
= \(\left[ \frac { k(k+1)(k+2) }{ 3 } \right] \)  (i)
For n = k + 1
P(k+1) = 1.2 + 2.3 + 3.4 +k(k + 1) + (k + 1) (k + 2)
=\(\frac { k(k+1)(k+2) }{ 3 } \) +(k+1)(k+2)   [using(i)] 
=(k+1)(k+2) \(\left[ \frac { k }{ 3 } +1 \right] \)
= (k+1)(k+2) \(\left[ \frac { k+3 }{ 3 } \right] \)
= \(\frac { (k+1)(k+2)(k+3) }{ 3 } \)
∴ 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) :\(1+2+3......n<\frac { 1 }{ 8 } { (2n+1) }^{ 2 }\)
Step II For n=1 , we have
\(n<\frac { 1 }{ 8 } {[ (2(1)+1)] }^{ 2 }\) \(\Rightarrow \) 1< \(\frac { 9 }{ 8 } \) which is true.
Thus, P(1) is true.
Step III Let us assume that P(k) is true.
i.e. P(k):\(1+2+3......k<\frac { 1 }{ 8 } { (2k+1) }^{ 2 }\)........(i)
Step IV   Now we shall prove the statement for n=k+1.For this, we have to show that
\(1+2+3......(k+1)<\frac { 1 }{ 8 } { [2(k+1)+1] }^{ 2 }\)
from Eq.(i) , We have  
\(1+2+3......k<\frac { 1 }{ 8 } { (2k+1) }^{ 2 }\)
\(1+2+3......k+(k+1)<\frac { 1 }{ 8 } { (2k+1) }^{ 2 }+(k+1)\)   [adding (k+1) on both sides]
\(=\frac { 1 }{ 8 } { [(2k+1) }^{ 2 }+8(k+1)]\)
\(\frac { 1 }{ 8 } { [4k }^{ 2 }+1+4k+8k+8]=\frac { 1 }{ 8 } { [4k }^{ 2 }+12k+9]\)
\(\frac { 1 }{ 8 } { (2k+3) }^{ 2 }=\frac { 1 }{ 8 } { [2(k+1)+1] }^{ 2 }\)
\(\Rightarrow \)  \(1+2+3......k+1<\frac { 1 }{ 8 } { [2(k+1)+1] }^{ 2 }\)
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):\({ 3 }^{ n }>{ 2 }^{ n }\)
Step II For =1,we have \({ 3 }^{ 1 }>{ 2 }^{ 1 }\)
\(\Rightarrow \) 3 > 2, Which is true.
Thus P(1) is true.
Step III Let us assume that P(k) is true.
i.e  P(k):\({ 3 }^{ k }>{ 2 }^{ k }\)
Step IV Now, we shall prove the statement for n=k+1.For this,we have to show \({ 3 }^{ k+1 }>{ 2 }^{ k+1 }\)
from Eq.(i) we have \({ 3 }^{ k }>{ 2 }^{ k }\)
\({ 3 }^{ k }.3>{ 2 }^{ k }.3\) [multiplying both sides by 3]
\(\Rightarrow \)  \({ 3 }^{ k+1 }>{ 2 }^{ k }.3\)
\({ 2 }^{ k }.3>{ 2 }^{ k }.3\) \(\Rightarrow \)  \({ 3 }^{ k }.3>{ 2 }^{ k }.2\) = \({ 2 }^{ k+1 }\)
Thus ,P(K+1) is true whenever P(k) is true. Hence, by principle of mathematical induction,P(n) is true for all \(n\in N\)

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

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):\({ k }^{ 3 }-7k+3=3\lambda \)
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): 2ⁿ > n
When n =1, 2¹ >1. Hence P(1) is true.
Assume that P(k) is true for any positive integer k, i.e.,
2^k > k ... (1)
We shall now prove that P(k +1) is true whenever P(k) is true.
Multiplying both sides of (1) by 2, we get
2. 2^k > 2k
i.e., \(2^{k+1}>2 k=k+k>k+1\)
Therefore, P(k + 1) is true when P(k) is true. Hence, by principle of mathematical induction, P(n) is true for every positive integer 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.

It is given that P(n) is true i.e.3ⁿ>n.
we have prove that P(n + 1) is true i.e. 3^{n + 1} >(n + 1)
Now 3ⁿ> n
⇒ 3·3ⁿ> 3n
⇒ 3ⁿ⁺¹>n+2n
⇒ 3ⁿ⁺¹>n+1         [∵ 2n> 1for all n∊N]
which shows that P(n+1) is true.

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

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):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)