# sum of binomial series

How do I solve sum of binomial series which is as follows: $$\frac{1}{3}\sum^{\infty}_{x=0}\begin{pmatrix}x\\y\end{pmatrix}\frac{1}{3}^x$$

I think it would be pretty easy to sum from $y=0$ to $x$, but I have no idea how to sum from $x=0$ to $x=\infty$.

Thanks,

• Is it $\frac {1^x}3$ or $\big(\frac 13\big)^x$ – Claude Leibovici May 3 '16 at 3:48

Remember binomial theorem: $$(p+q)^x = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} q^y p^{x-y}$$ With $p = 1/3$: $$(1/3+q)^x = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} q^y \left( \frac{1}{3} \right)^{x-y} = \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} 3^y q^y \left( \frac{1}{3} \right)^{x}$$ Now summing over $x$ on both sides and rearranging some things in the sum gives: $$\sum_{x = 0}^\infty (1/3 + q)^x = \sum_{x} \sum_{y} \begin{pmatrix}x\\y\end{pmatrix} 3^y q^y \left( \frac{1}{3} \right)^{x} \\ =\sum_{y} q^y 3^y \; \sum_{x} \begin{pmatrix}x\\y\end{pmatrix} \left( \frac{1}{3} \right)^{x}$$ So all you have to do to find your sum is to find the coefficient of $q^y$ term in $\sum_{x = 0}^\infty (1/3 + q)^x$, and divide by $3^y$. The sum is easily computed by geometric series: $$\sum_{x = 0}^\infty (1/3 + q)^x = \frac{1}{2/3 - q} \\ = \frac{3}{2} \sum_y (3 q /2)^y$$ Where in the last line I used the geometric series expansion of $1/(1 - x)$ The coefficient of $q^y$ is $(3/2)^{y+1}$, dividing by $3^y$ gives $3/ 2^{y+1}$.