For xR, x-1, if (1+x)^2016+x(1+x)^2015+x^2(1+x)^2014++x^2016=_i=0^2016a_ix^i, then a_17 is…

Question

For $x\in\mathbb{R}, x
e-1$, if $(1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\cdots+x^{2016}=\displaystyle\sum_{i=0}^{2016}a_ix^i$, then $a_{17}$ is equal to:

Accepted Answer

The left side is a geometric series in the ratio $\dfrac{x}{1+x}$ with 2017 terms and first term $(1+x)^{2016}$, so it sums to
$$\frac{(1+x)^{2017}-x^{2017}}{(1+x)-x}=(1+x)^{2017}-x^{2017}.$$

The coefficient of $x^{17}$ in $(1+x)^{2017}$ is $\binom{2017}{17}$, and $x^{2017}$ contributes nothing to $x^{17}$ (since $2017\ne17$). So
$$a_{17}=\binom{2017}{17}=\frac{2017!}{17!\,2000!}.$$

Solutions · 1