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  • Let the coefficients of three consecutive terms in the binomial expansion of <img alt="\mathrm{\left ( 1+2x \right )^{n}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cl

Let the coefficients of three consecutive terms in the binomial expansion of \mathrm{\left ( 1+2x \right )^{n}} be in the ratio \mathrm{2: 5: 8}. Then the coefficient of the term, which is in the middle of these three terms, is

Option: 1

1120


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let \mathrm{r}+1, \mathrm{r}+2$ and $\mathrm{r}+3 be three consecutive terms

\begin{aligned} & \frac{{ }^n C_r 2^r}{{ }^n C_{r+1} 2^{r+1}}=\frac{2}{5} \\ \end{aligned}

\begin{aligned} & \Rightarrow \frac{r+1}{n-r}=\frac{4}{5} \end{aligned}..............(1)
Also,
\begin{aligned} & \frac{{ }^n C_{r+1} 2^{\mathrm{r}+1}}{{ }^4 C_{r+2} 2^{r+2}}=\frac{5}{8} \\ \end{aligned}

\begin{aligned} & \Rightarrow \frac{r+2}{n-r-1}=\frac{5}{4} \end{aligned}.................(2)

on solving (1) \& (2), we get

\mathrm{n}=8, \mathrm{r}=3
Here \mathrm{n=8} (even)

\mathrm{ \text { middle term }=r+2=3+2=5}

\mathrm{ \text { coefficient of } T_5={ }^8 \mathrm{C}_4 2^4=70(16)=1120}

Posted by

Shailly goel

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