Question

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

• A. 64
• B. 2187
• C. 243
• D. 729

Answer 1

The correct option is D 729

Clearly, number of terms in the expansion of
${(1-\frac{2}{x}+\frac{4}{x^2})}^{n}is\frac{(n+2)(n+1)}{2}or{}^{n+2}{C}_2$
[assuming $\frac{1}{x}$ and $\frac{1}{x^2}$ distinct]
$\therefore \frac{(n+2)(n+1)}{2}=28$
$\Rightarrow$ (n + 2)(n + 1) = 56 =( 6 + 1) (6 + 2) $\Rightarrow$ n = 6
Hence, sum of coefficients $=(1-2+4{)}^6=3^6=729$