If the number of terms in the expansion | vedclass

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Question

Clearly, number of terms in the expansion of

$\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}$ is $\frac{(n+2)(n+1)}{2}$ or $^{n+2} C_{2}$

[assum

Vedclass · Accepted Answer

$729$

Vedclass · Answer

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

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

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