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 expansion, is :

• A. 2187
• B. 243
• C. 729
• D. 64

Answer 1

The correct option is C

729

Number of terms is 2n + 1 which is odd but it is given 28. If we take $(x+y+z{)}^n$ then number of terms is ${}^{n+2}{C}_2=28$ Hence n = 6

${(1-\frac{2}{x}+\frac{4}{x^2})}^6$ $=a_0+a_{1}x$ $+a_2{x}^2+$ $\cdots \cdots +a_6{x}^6$

Sum of coefficients can be obtained by x = 1

$(1-2+4{)}^6=3^6=729$

So according to what the examiner is trying to ask option 3 can be correct.