If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1 + x)^{10}$, then $$\sum\limits_{r = 1}^{10} {{r^3}{{\left( {{{{a_r}} \over {{a_{r - 1}}}}} \right)}^2}} $$ is equal to
Solution
$$
\begin{aligned}
& \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{10-\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}} \\\\
& \Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{10} \mathrm{C}_{\mathrm{r}-1}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{11-\mathrm{r}}{\mathrm{r}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}(11-\mathrm{r})^2 \\\\
& =\sum_{\mathrm{r}=1}^{10}\left(121 \mathrm{r}+\mathrm{r}^3-22 \mathrm{r}^2\right)=1210
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
This question is part of PrepWiser's free JEE Main question bank. 193 more solved questions on Binomial Theorem are available — start with the harder ones if your accuracy is >70%.