<s>[INST] 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 $A_9 = 1, A_0 = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5$.
 [/INST] Let $P$ be your polynomial : 
$$P(x)=(x+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)=\sum_{k=0}^9 A_kx^k$$
Then 
$$\sum_{k=0}^9 A_k = \sum_{k=0}^9 A_k.1^k = P(1) = 2\times3\times4\times5\times6 = 720$$
And, as
$$\sum_{k=0}^9 (-1)^kA_k = P(-1) = 0$$
you can, by adding, find :
$$2\sum_{k=0}^4 A_{2k} = P(1)+P(-1)=720$$
so that 
$$\sum_{k=0}^4 A_{2k} = 360$$
</s>