Sum of coefficients of even power of x in expansion of {(1 + x + {x^2} + {x^3})^5} is

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7.Binomial Theorem

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The sum of the coefficients of even power of $x$ in the expansion of ${(1 + x + {x^2} + {x^3})^5}$ is

A

$256$

B

$128$

C

$512$

D

$64$

Solution

(c) ${(1 + x + {x^2} + {x^3})^5} = {(1 + x)^5}{(1 + {x^2})^5}$

$ = (1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5})$$ \times (1 + 5{x^2} + 10{x^4} + 10{x^6} + 5{x^8} + {x^{10}})$

Therefore the required sum of coefficients

$ = (1 + 10 + 5){.2^5} = 16 \times 32 = 512$

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

Standard 11

Mathematics

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