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) : 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

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]