For all $$n \in N, 2^{4n} − 15n − 1$$​ is divisible by:

Correct option is A

Given:
​$$2^{4n} − 15n − 1$$​​
Formula used:
Binomial expansion:
​$$(1 + x)^{n} = 1 + nx + \frac{n(n-1)}{2}x^{2} + \cdots$$​​
Solution:
​$$2^{4n} = (2^{4})^{n} = 16^{n}$$​​
Write:
​$$16 = 1 + 15$$​​
So,
​$$16^{n} = (1 + 15)^{n}$$​​
Using binomial expansion:
​$$(1 + 15)^{n}= 1 + 15n + \frac{n(n-1)}{2} \cdot 15^{2} + \cdots$$​​
Now subtract (15n + 1):
​$$16^{n} − 15n − 1$$​​
​$$= \frac{n(n-1)}{2} \cdot 15^{2} + \cdots$$​​
Each remaining term contains the factor $$15^{2} = 225.$$​​
Hence,
$$2^{4n} − 15n − 1$$​ is divisible by 225 for all n \in N.
The correct answer is (a) 225.