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

JEE Main · 2016 · 03 Apr 2016 · Offline · Binomial Theorem

If the number of terms in the expansion of

(1 - \frac{2}{x} + \frac{4}{x ^{2}})^{n}, x \neq = 0

, is 28, then the sum of the coefficients of all the terms in this expansion, is:

Answer: (C)

Read

(1 - \frac{2}{x} + \frac{4}{x ^{2}})^{n}

as a trinomial raised to the power

n

. Counting terms the way the paper intends — treating each choice among the three parts separately, without merging equal powers of

\frac{1}{x}

— the number of terms is

(2 n + 2)

. Setting

(2 n + 2) = 28

gives

(n + 1) (n + 2) = 56

, so

n = 6

. The sum of all the coefficients comes from substituting

x = 1

, which gives

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

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option