Let left(2 x ^{2}+3 x +4right)^{10}=sum limit | vedclass

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Question

Given $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{ r =0}^{20} a _{ r } x ^{ r }$

replace x by $\frac{2}{x}$ in above identity :-

$\frac{2^{10}\left(2

Vedclass · Accepted Answer

$8$

Vedclass · Answer

Given $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{ r =0}^{20} a _{ r } x ^{ r }$

replace x by $\frac{2}{x}$ in above identity :-

$\frac{2^{10}\left(2 x ^{2}+3 x +4\right)^{10}}{ x ^{20}}=\sum_{ r =0}^{20} \frac{ a _{ r } 2^{ r }}{ x ^{ r }}$

$\Rightarrow 2^{10} \sum_{ r =0}^{20} a _{ r } x ^{ r }=\sum_{ r =0}^{20} a _{ r } 2^{ r } x ^{(20- r )}($ from

now, comparing coefficient of $x^{7}$ from both sides

(take $r=7$ in L.H.S. $\& r=13$ in $R .$ H.S. $)$

$2^{10} a_{7}=a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}}=2^{3}=8$

Let $\left(2 x ^{2}+3 x +4\right)^{10}=\sum \limits_{ r =0}^{20} a _{ r } x ^{ r } \cdot$ Then $\frac{ a _{7}}{ a _{13}}$ is equal to

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