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Q. 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 MainJEE Main 2016Binomial Theorem

Solution:

Number of terms $ = \frac{(n +1)(n+2)}{2} = 28 $
$\Rightarrow n = 6$
$\therefore \, \, a_0 + \frac{a_1}{x} + \frac{a_2}{x^2} + .... + \frac{a_2n}{x^{2n}} = \left( 1 - \frac{2}{x} + \frac{4}{x^2} \right)^n$
Put $x = 1, n = 6 , a_0 + a_1 + a_2 + ... + a_{2n} = 3^6 = 729$