If a_r is the coefficient of x^10-r in the Binomial expansion of (1+x)^10, then _r=1^10r^3…

Question

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\displaystyle\sum_{r=1}^{10}r^3\left(\dfrac{a_r}{a_{r-1}}\right)^2$ is equal to

Accepted Answer

$$
\begin{aligned}
& \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{10-\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}} \\
& \Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{10} \mathrm{C}_{\mathrm{r}-1}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{11-\mathrm{r}}{\mathrm{r}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}(11-\mathrm{r})^2 \\
& =\sum_{\mathrm{r}=1}^{10}\left(121 \mathrm{r}+\mathrm{r}^3-22 \mathrm{r}^2\right)=1210
\end{aligned}
$$

Solutions · 1