Let $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{r=0}^{20} a_{r} x^{r}$. Then $\frac{a_{7}}{a_{13}}$ is equal to

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Binomial Theorem

Let

(2 x^{2} + 3 x + 4)^{10} = \sum_{r = 0}^{20} a_{r} x^{r}

. Then

\frac{a _{7}}{a _{13}}

is equal to

Answer: 8

Given

(2 x^{2} + 3 x + 4)^{10} = \sum_{r = 0}^{20} a_{r} x^{r}

, replace

x

\frac{2}{x}

: the left side becomes

(\frac{8}{x ^{2}} + \frac{6}{x} + 4)^{10} = \frac{2 ^{10}}{x ^{20}} (2 x^{2} + 3 x + 4)^{10}

, so

\frac{2 ^{10}}{x ^{20}} \sum_{r} a_{r} x^{r} = \sum_{r} a_{r} \frac{2 ^{r}}{x ^{r}}

. Multiplying by

x^{20}

gives

2^{10} \sum_{r} a_{r} x^{r} = \sum_{r} a_{r} 2^{r} x^{20 - r}

. Comparing the coefficient of

x^{7}

(which is

x^{20 - r}

with

r = 13

on the right) gives

2^{10} a_{7} = 2^{13} a_{13}

, so

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

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Integer answer