Question

If the number of terms in the expansion of ${\left(1-\frac{2}{x}+\frac{4}{x^2}\right)}^n,x\ne 0$, is $28$, then the sum of coefficients of all the terms in this expansion, is:

• A. $2187$
• B. $243$
• C. $729$
• D. $64$

Answer 1

The correct option is C $729$

No. of terms in expansion of ${\left(1-\frac{2}{x}+\frac{4}{x^2}\right)}^n$ is

${}^{n+3-1}{c}_{3-1}=28$

${}^{n+2}{c}_2=28$

$\frac{n+2!}{2!n!}=28$

$\left(n+2\right)\left(n+1\right)=8\times 7$

$n+2=8$

$n=6$

${\left(1-\frac{2}{x}+\frac{4}{x^2}\right)}^6={\left(1-2+4\right)}^6$

$={\left(3\right)}^6=729$

Hence,

option $\left(C\right)$ is correct answer.