The coefficient of x^4 in the expansion of (1 | vedclass

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(C) We have the expression $(1+x-x^2-x^3)^{11}$.Factorizing the expression inside the bracket:$(1+x-x^2-x^3) = 1(1+x) - x^2(1+x) = (1-x^2)(1+x) = (1-x

Vedclass · Accepted Answer

$-220$

Vedclass · Answer

(C) We have the expression $(1+x-x^2-x^3)^{11}$.Factorizing the expression inside the bracket:$(1+x-x^2-x^3) = 1(1+x) - x^2(1+x) = (1-x^2)(1+x) = (1-x)(1+x)(1+x) = (1-x)(1+x)^2$.Thus,the expression becomes $((1-x)(1+x)^2)^{11} = (1-x)^{11}(1+x)^{22}$.We need the coefficient of $x^4$ in the expansion of $(1-x)^{11}(1+x)^{22}$.$(1-x)^{11} = \sum_{r=0}^{11} (-1)^r {}^{11}C_r x^r$ and $(1+x)^{22} = \sum_{k=0}^{22} {}^{22}C_k x^k$.The coefficient of $x^4$ is given by $\sum_{r=0}^4 (-1)^r {}^{11}C_r \cdot {}^{22}C_{4-r}$.$= {}^{11}C_0 \cdot {}^{22}C_4 - {}^{11}C_1 \cdot {}^{22}C_3 + {}^{11}C_2 \cdot {}^{22}C_2 - {}^{11}C_3 \cdot {}^{22}C_1 + {}^{11}C_4 \cdot {}^{22}C_0$.$= 1 \cdot 7315 - 11 \cdot 1540 + 55 \cdot 231 - 165 \cdot 22 + 330 \cdot 1$.$= 7315 - 16940 + 12705 - 3630 + 330 = -220$.

The coefficient of $x^4$ in the expansion of $(1+x-x^2-x^3)^{11}$ is

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