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)

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]

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 \(P(k) : \frac { { k }^{ 7 } }{ 7 } +\frac { { k }^{ 5 } }{ 5 } +\frac { 2 }{ 3 } k^{ 3 }-\frac { k }{ 105 } \lambda \in I\)
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

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): 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): \(1+3+3^{ 2 }+.....+3^{ n-1 }=\frac { 3^{ n }-1 }{ 2 } \)
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 \(n \in N\)

P(n) =1³+2³+3³+......+n³= \(\left[ \frac { n(n+1) }{ 2 } \right] ^{ 2 }\)
For n =1
P(1) = 1=\(\left[ \frac { 1(1+1) }{ 2 } \right] ^{ 2 }\)
⇒ 1=1
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = 1³ + 2³ + 3³ +.....+ k³
=  \(\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }\)          ..(i)
For n = k + 1
P(k + 1) =1³+ 2³+ 3³....+k³+(k + 1)³
=\(\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }\)+(k+1)³  [Using (i)]
= (k+1)² \(\left[ \frac { { k }^{ 2 } }{ 4 } +k+1 \right] \)
= (k+1)² \(\left[ \frac { { k }^{ 2 }+4k+4 }{ 4 } \right] \)
=  \(\frac { (k+1)^{ 2 }+(k+2)^{ 2 } }{ 4 } =\left[ \frac { (k+1)(k+2) }{ 2 } \right] ^{ 2 }\)
∴ P(k + 1) is true.
Thus P(k) is true ⇒ (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all n ∊ N.