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Let (2x^{2}+3x+4)^{10}=\sum_{r=0}^{20}a_{r}x^{r}. Then \frac{a_{7}}{a_{13}} is equal to ________
Option: 1 2
Option: 2 4
Option: 3 6
Option: 4 8

Answers (1)

best_answer

\text { Given }\left(2 x^{2}+3 x+4\right)^{10}=\sum_{r=0}^{20} a_{r} x^{r}\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)

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

\Rightarrow 2^{10} \sum^{20} \mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}=\sum^{20} \mathrm{a}_{\mathrm{r}} 2^{\mathrm{r}} \mathrm{x}^{(20-\mathrm{r})} \text { (from \;(1))}

\text { now, comparing coefficient of } \mathbf{x}^{7} \text { from both sides }

(take r = 7 in L.H.S. & r = 13 in R.H.S.)

2^{10} \mathrm{a}_{7}=\mathrm{a}_{13} 2^{13} \Rightarrow \frac{\mathrm{a}_{7}}{\mathrm{a}_{13}}=2^{3}=8

Posted by

himanshu.meshram

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