First slide

Binomial theorem for positive integral Index

Question

If the coefficients of $x^{7}$ and $x^{8}$ in ${(2 + \frac{x}{3})}^{n}$ are equal, then value of $n$ is

Moderate

A

$56$
B

$55$
C

$47$
D

$19$

Solution

$T_{r + 1},$ the $(r + 1) t h$ term in the expansion of ${(2 + \frac{x}{3})}^{n}$ is given by
$T_{r + 1} =^{n} C_{r} 2^{n - r} {(\frac{x}{3})}^{r} =^{n} C_{r} (\frac{2^{n - r}}{3^{r}}) x^{r}$
According to the given condition
$\begin{array}{l} ^{n} C_{7} (\frac{2^{n - 7}}{3^{7}}) =^{n} C_{8} (\frac{2^{n - 8}}{3^{8}}) \Rightarrow \frac{^{n} C_{7}}{^{n} C_{8}} = \frac{2^{n - 8}}{3^{8}} \cdot \frac{3^{7}}{2^{n - 7}} \\ \Rightarrow \frac{n!}{7! (n - 7)!} \frac{8! (n - 8)!}{n!} = \frac{1}{6} \Rightarrow \frac{8}{n - 7} = \frac{1}{6} \\ \Rightarrow 48 = n - 7 \Rightarrow n = 55 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App