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.

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, \(n\ge 3\).

Here, the given statement is true for all n \(\in\) N
it is true for n=1 and n=2
For n=1, P(1) : 3.5+2⁴=3x125x16
=375+16=391
and for n=2, P(2) ; 3.5⁵+2⁷
=3x3125+128
=9375+128=9503
Now, the HCF of 391 and 9503 is 17. So,
3.5²ⁿ⁺¹+2³ⁿ⁺¹ is divisible by 17. Hence, \(\lambda \) is 17.

Here, the given statement is true for all n \(\in\) N, therefore it is true for n=1 also
So, we have P(1) : 49+1++K is divisible by 64,
i.e. 65+k is divisible be 64
Clearly, k should be -1  [\(\therefore \)65-1=64 is divisible by 64]

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.
\(P(n):\quad 1+2+3+.....+n=\frac { n(n+1) }{ 2 } \)
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.

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 \(n\in N\)

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.

Let P(n) be the given statement,
i.e. P(n) : (ab)ⁿ=aⁿ  b ⁿ
We note that P(n) is true For n=1, (ab)¹ =ab=a¹ b¹
Let P(k) be true, i.e.,
Then, we have (ab)^k =a^k b^k  ..(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Now, we have
(ab)^{k + 1} = (ab)^k (ab)
=( a^k b^k ) (ab)       [using Eq.(i)]
 =(a^k .a¹) (b^k.b¹)=a^k+1.b^k+1 
Thus, P(K+1) is also 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): \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 }=\left[ \frac { n(n+1) }{ 2 } \right] ^{ 2 }\)
Step II  For n=1, we have, LHS=1³=1
and RHS= \(\left[ \frac { 1(1+1) }{ 2 } \right] ^{ 3 }=\left( \frac { 1.2 }{ 2 } \right) ^{ 3 }=1\)
LHS=RHS;     P(1) is true
Step III  Let us assume that P(n) is true for n=k. Then we have
P(k): \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.....+{ k }^{ 3 }=\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }...(i)\)
Step IV  Now, we shall prove the statement for n=k+1 For this we have to show
1³+2³+3³+.....+k³+(k+1)³
=\(\left[ \frac { (k+1)(k+1+1) }{ 2 } \right] ^{ 2 }\)
Now, LHS=1³+2³+3³+.....+k³+(k+1)³
= \(\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }+(k+1)^{ 3 }\)       [from Eq.(i)]
\(=(k+1)^{ 2 }\left[ \frac { k^{ 2 } }{ 4 } +(k+1) \right] \)
\(\\ =(k+1)^{ 2 }\left[ \frac { k^{ 2 }+4(k+1) }{ 4 } \right] =\frac { (k+1)^{ 2 }\left[ k^{ 2 }+4k+4 \right] }{ 4 } \)
\(=\frac { (k+1)^{ 2 }(k+2)^{ 2 } }{ 4 } =RHS\)
So, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n\in N\)