Question

If the number of term 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 term in this expansion, is:

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

Answer 1

The correct option is C $729$

Number of terms $=\frac{(n+1)(n+2)}{2}=28$

$\Rightarrow$ $(n+1)(n+2)=56$

$\Rightarrow$ $n^2+2n+n+2=56$

$\Rightarrow$ $n^2+3n-54=0$

$\Rightarrow$ $n^2+9n-6n-54=0$

$\Rightarrow$ $(n+9)(n-6)=0$

$\therefore$ $n=6$ since $n=-9$ is not possible.

$a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+....+\frac{a_{2n}}{x^{2n}}={(1-\frac{2}{x}+\frac{4}{x^2})}^n$

Substitute $x=1$ and $n=6$

$a_0+a_1+a_2+...+a_{2n}=3^6=729$