7.Binomial Theorem
hard

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

A

$4895$

B

$1210$

C

$5445$

D

$3025$

(JEE MAIN-2023)

Solution

$a _{ r }={ }^{10} C _{10- r }={ }^{10} C _{ r }$

$\Rightarrow \sum \limits_{ r =1}^{10} r ^3\left(\frac{{ }^{10} C _{ r }}{{ }^{10} C _{ r -1}}\right)^2=\sum \limits_{ r =1}^{10} r ^3\left(\frac{11- r }{ r }\right)^2=\sum \limits_{ r =1}^{10} r (11- r )^2$

$=\sum \limits_{ r =1}^{10}\left(121 r + r ^3-22 r ^2\right)=1210$

Standard 11
Mathematics

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