The coefficient of x 4 of in the expansion ( 1 + 5 x + 9 x 2 + …… ∞ ) ( 1 + x 2 ) 11 is

1 + 5 x + 9 x 2 + ⋯ = ∑ k = 0 ∞ ( 4 k + 1 ) x k and ( 1 + x 2 ) 11 = ∑ j = 0 11 11 C j x 2 j Coefficient of x 4 = ( 4.4 + 1 ) . 11 C 0 + ( 4.2 + 1 ) . 11 C 1 + ( 4.0 + 1 ) . 11 C 2 = 17 + 99 + 55 = 171

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MathematicsBinomial TheoremBinomial Theorem for Negative IndexMedium2 minQB

MathematicsMediumsingle choice

The coefficient of $x^{4}$ of in the expansion $(1 + 5 x + 9 x^{2} + \dots\dots \infty) (1 + x^{2})^{11}$ is

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$1 + 5 x + 9 x^{2} + \dots = \sum_{k = 0}^{\infty} (4 k + 1) x^{k}$ and $(1 + x^{2})^{11} = \sum_{j = 0}^{11}^{11} C_{j} x^{2 j}$ Coefficient of $x^{4} = (4.4 + 1) .^{11} C_{0} + (4.2 + 1) .^{11} C_{1} + (4.0 + 1) .^{11} C_{2}$ $= 17 + 99 + 55 = 171$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Binomial Theorem for Negative Index

⏱ 2mℹ️ Source: QB

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