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The sum of the coefficients of even power of x in the expansion of \left(1+x+x^2+x^3\right)^5 is

Option: 1

256


Option: 2

128


Option: 3

512


Option: 4

64


Answers (1)

best_answer

\begin{aligned} \left(1+x+x^2+x^3\right)^5 & =(1+x)^5\left(1+x^2\right)^5 \\ & =\left(1+5 x+10 x^2+10 x^3+5 x^4+x^5\right) \ \times\left(1+5 x^2+10 x^4+10 x^6+5 x^8+x^{10}\right) \end{aligned}

Therefore the required sum of coefficients

=(1+10+5) \cdot 2^5=16 \times 32=512

Note : 2^n=2^5= Sum of all the binomial coefficients in the 2nd bracket in which all the powers of x are even.

 

Posted by

Sanket Gandhi

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