Calculate sum of coefficients of polynomial Let $$(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5) = \sum_{k=0}^{9} (A_k \cdot x^k)$$
Compute:


*

*$$\displaystyle \sum_{k=0}^{9} A_k$$

*$$\displaystyle \sum_{k=0}^{4} A_{2k}$$
I tried to figure out from Viete's Sums how to rewrite this but I can't find the coefficients for all powers of $x$. All I know is $A_9 = 1, A_0 = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5$.

Well, since
$$\sum_{k=0}^{9} A_kx^k=(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5)$$
we can just set $x=1$ to see
$$\sum_{k=0}^{9} A_k=(1 + 1)(1^2 + 2)(1^2 + 3)(1^2 + 4)(1^2 + 5)=720$$

Try setting $x=-1$; can you now find $\sum_{k=0}^{4} A_{2k}$?
