Value of $a_2+a_6+a_{10}+\cdots+a_{42}$

Question

If $(1+x+x^2+\cdots+x^9)^4(x+x^2+x^3+\cdots+x^9)=\sum_{r=1}^{45} a_rx^r ,$ then what is the value of $a_2+a_6+a_{10}+\cdots+a_{42}$

Answer 1

Calling your expression $f(x)$, we have $$ a_2 + a_6 + \ldots + a_{42} = \dfrac{f(1) + f(-1) - f(i) - f(-i)}{4} $$ Now $f(1) = 9 \cdot 10^4$, $f(-1) = 0$, $f(i) = -4 i$ and $f(-i) = 4i$, so this is $(9 \cdot 10^4)/4 = 22500$.