Q: How to find the sum of the coefficients of the product of two polynomials in $R[x]$? The following question was asked: "Consider the formal power series $\sum_{n\geq0}x^{3^{2n}-1}$ and $\sum_{n\geq0}x^{2^{2n}}$. Write their product as an expression in R[[x]] and find the sum of the coefficients". I have found that the two formal power series can be written as $\prod_{n\geq1}\left(1-x^{3^n}\right)$ and $\prod_{n\geq1}\left(1-x^{2^n}\right)$ in the ring of power series. Can anyone help me to find the sum of the coefficients? A: Let $A(x) = \prod_{n\geq 1} (1-x^n)^{3^{2n}-1}$ and $B(x) = \prod_{n\geq 1} (1-x^n)^{2^{2n}}$. It's easily verified that $A(x^3) = B(x^2)$. Since $A(x) = \sum_{i=1}^\infty \operatorname{coeff}(A;i) x^i$ and $B(x) = \sum_{i=1}^\infty \operatorname{coeff}(B;i) x^i$, you obtain the claim by comparing the coefficient of $x^i$ on both sides. A: You can do it here. A: If you have no idea of abstract algebra, you can proceed this way: $$\begin{align*}&(1+x+x^2+...+x^{3^n-1})+(1+x+x^2+...+x^{2^n})\\ =&1+x+x^2+...+x^{3^n-1}+x+x^2+...+x^{2^n-1}+x^{3^n+1}+x^{3^n}+x^{3^n-2}+...\\ =&1+x+x^2+...+x^{3^n-1}+x^{3^n}+x^{2^n}+...+x^{3^{n-1}}+x+x^2+...+x^{2^n-1}+x^{2^{n+1}}+...\\ =&(1+x+x^2+...+x^{2^{n-1}})+(1+x+x^2+...+x^{2^n})\\ =&1+x+x^2+...+x^{2^n-1}+1+x+x^2+...+x^{2^n-1}\\ =&1+x+x^2+...+x^{2^n-1}+x+x^2+...+x^{2^n-1}\\ =&2(1+x+x^2+...+x^{2^n-1}) \end{align*}$$ So, what we have is $A(x)=1+x+x^2+...+x^{2^n-1}$, and $B(x)=2A(x)$. Now, you can apply any method you know to get the sum of the coefficients of $B(x)$ (without knowing algebra, that is). EDIT You can see here that, if you have two polynomials $A(x)$ and $B(x)$ defined over $\mathbb Z_m$, and such that $B(x)=2A(x)$, you have $\operatorname{coeff}(B,i)=\operatorname{coeff}(A,i)+2\operatorname{coeff}(A,i-1)$. In your case, $\operatorname{coeff}(B,i)=\operatorname{coeff}(A,i)+2\operatorname{coeff}(A,i-1)$. Now, to calculate the sum of the coefficients of $A(x)$, we have $\sum_{i=0}^n \operatorname{coeff}(A,i)=A(0)+A(1)+...+A(n)=1+1+...+1=n+1$. You can check the last identity as follows: $A(x)=1+x+x^2+...+x^{m-1}$ is a polynomial defined over $\mathbb Z_m$, and $A(0)=1$, $A(1)=1$, ..., $A(m-1)=1$. I don't know how to get the sum of the coefficients of $B(x)$ using only the formal manipulation of polynomials, but here is how it could be done. The coefficient of $x^{3^k-1}$ in $B(x)$ is just the coefficient of $x^{2^k}$ in $A(x)$, that is, $\operatorname{coeff}(B,3^k-1)=\operatorname{coeff}(A,2^k)$, or $\sum_{j=0}^{2^k}\operatorname{coeff}(B,j)=\sum_{j=0}^{2^k}\operatorname{coeff}(A,j)$. We can evaluate the left and right parts for any $k$: $$\begin{align*}& \sum_{j=0}^{2^k}\operatorname{coeff}(B,j)\\ =&\sum_{j=0}^{2^k}\left[\operatorname{coeff}(A,j)+2\operatorname{coeff}(A,j-1)\right]\\ =&\operatorname{coeff}(A,0)+2\operatorname{coeff}(A,0)+\operatorname{coeff}(A,1)+2\operatorname{coeff}(A,1)+...+\operatorname{coeff}(A,2^k)+\operatorname{coeff}(A,2^k)\\ &\qquad\qquad-2\operatorname{coeff}(A,0)-2\operatorname{coeff}(A,1)-...-2\operatorname{coeff}(A,2^k)\\ =&\operatorname{coeff}(A,0)+2\operatorname{coeff}(A,1)+2\operatorname{coeff}(A,2)+...+2\operatorname{coeff}(A,2^k)+2\operatorname{coeff}(A,2^k+1)\\ &\qquad\qquad+\operatorname{coeff}(A,3^k-1)+\operatorname{coeff}(A,3^k-2)\\ =&\operatorname{coeff}(A,0)+2\operatorname{coeff}(A,1)+2\operatorname{coeff}(A,2)+...+2\operatorname{coeff}(A,2^k)+2\operatorname{coeff}(A,2^k+1)\\ &\qquad\qquad+\operatorname{coeff}(A,3^k-1)+\operatorname{coeff}(A,3^k-2)\\ =&\sum_{j=0}^{2^{k+1}}\operatorname{coeff}(A,j)\\ \end{align*}$$ Since $k$ is any integer, you can get $\sum_{j=0}^{2^{2k}}\operatorname{coeff}(B,j)$ by induction. Now, we just have to evaluate $\sum_{j=0}^{2^{2k}}\operatorname{coeff}(A,j)$. You can prove by induction that $\operatorname{coeff}(A,j)=\frac{1}{2}(1+(-1)^j)$. In that case, we will have: $$\begin{align*}&\sum_{j=0}^{2^{2k}}\operatorname{coeff}(A,j)\\ =&\frac{1}{2}\sum_{j=0}^{2^{2k}}(1+(-1)^j)\\ =&\frac{1}{2}\left(\frac{1}{2}+(-1)^{2k}+\frac{1}{2}(-1)^{2k}\right)\\ =&\frac{1}{2}\cdot2\\ =&1 \end{align*}$$ Now, you can calculate the sum of the coefficients of $B(x)$ with no difficulties: $$\begin{align*}& \sum_{j=