What's the sum?

Let $(1+x+x^2+x^3)^5=\sum_{i=0}^{15} a_i x^i$ , where $a_i$ are the coefficients. It is required to find $\sum_{i=0}^{15} a_i$

Substituting $x=1$ in the equation

$\sum_{i=0}^{15} a_i = (1+1+1+1)^5=4^5=\boxed{1024}$

Nice Problem, and this is my solution :D

Let the expansion is of the form:- $(1+x+x^{2}+x^{3})^{5} = a_{0} + a_{1}x+a_{2}x^{2} + \ldots + a_{15}x^{15}$ The expansion is taken for 16 terms only because the highest power of x that can reach is 15.

It is to be remembered that the above written expression is an identity and not an equation.

Hence the above expansion holds true for any value of $x$ . So if we replace $x$ by $1$ we'll get:- $4^{5} = a_{0} + a_{1}+a_{2}+\ldots + a_{15}$ Hence we get our sum of coefficients.

4 numbers(N) to the 5th power means: N x N x N x N x N, which is 4^5, thus the answer is 1024.