Most Important Selected Qs for JEE AdvancedMathematicsBinomial Theorem

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

Options

A256
B128
C512
D64

Correct answer

C. 512

Step-by-step solution

(1+x+x^2+x^3 )^5=a₀+a₁ x+a₂ x^2+a₃ x^3+a₄ x^4+ . .+a₁₅ x¹⁵ Put x=1 and x=-1 and add aligned & 4^5=a₀+a₁+a₂+a₃+a₄+ . .+a¹⁵ & 0=a₀-a₁+a₂-a₃+a₄- . .-a¹⁵ aligned aligned & 4^5=2 (a₀+a₂+a₄+ +a₁₄ ) & a₀+a₂+a₄+ +a₁₄=512 aligned

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