In the expansion of ${(\frac{1}{2} + \frac{2 x}{3})}^{n}$ when $x = - \frac{1}{2}$ it is known that $3^{r d}$ term is the greatest term Find the sum of all possible integral values of n

Open in App

Solution

Substituting $x = - \frac{1}{2}$
Hence
$(\frac{1}{2} - \frac{1}{3})^{n}$
${\frac{1}{2}}^{n} (1 - \frac{2}{3})^{n}$
For numerically greatest term.
$r = \frac{(1 + n) | x |}{1 + | x |}$
$r = 2$
Hence
$2 + 2 | x | = | x | (1 + n)$
$2 + \frac{4}{3} = \frac{2}{3} (1 + n)$
$\frac{10}{3} = \frac{2}{3} (1 + n)$
$5 = 1 + n$
$n = 4$
Now $n$ ranges from $4$ to $6$
Hence
$4 + 5 + 6 = 15$

Suggest Corrections

Similar questions

Q. In the expansion of

{(\frac{1}{2} + \frac{2 x}{3})}^{n}

when

x = - \frac{1}{2}

, it is known that 3rd term is the greatest term. Find possible integral values of

n

Q. In the expansion of

(\frac{1}{2} + \frac{2 x}{3})

, where

x = - \frac{1}{2}

, it is known that

3^{r d}

term is the greatest term. Find the possible integral values of

n

Q. In the expansion of

{(\frac{3}{2} + \frac{7}{3})}^{n}

when

x = \frac{1}{2}

if is known that

6^{t h}

term is the greatest term the find the possible integral value of n.

If $6^{t h}$ term in the expansion of $(\frac{3}{2} + \frac{x}{3})^{n}$ is the numerically greatest term when x=3, then find the sum of all possible values of n

Q. If the

6^{t h}

term in the expansion of

{(\frac{3}{2} + \frac{x}{3})}^{n}

when

x = 3

is numerically greatest then the possible integral values of n can be

Join BYJU'S Learning Program

In the expansion of (12+2x3)n when x=−12 it is known that 3rd term is the greatest term Find the sum of all possible integral values of n

Substituting x=−12Hence(12−13)n12n(1−23)nFor numerically greatest term.r=(1+n)|x|1+|x|r=2Hence2+2|x|=|x|(1+n)2+43=23(1+n)103=23(1+n)5=1+nn=4Now n ranges from 4 to 6Hence4+5+6=15

In the expansion of ${(\frac{1}{2} + \frac{2 x}{3})}^{n}$ when $x = - \frac{1}{2}$ it is known that $3^{r d}$ term is the greatest term Find the sum of all possible integral values of n