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7.Binomial Theorem
hard
$(1+x)^{10}$ ના દ્વિપદી વિસ્તરણમાં $x^{10-r}$ નો સણગુણક જો $a_r$ હોય., તો $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2=...............$
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
