Sum of the coefficients of x^4 and x^6 in the | vedclass

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(C) The expansion is given by $(1 + (x - x^2))^6$. Using the binomial theorem,$(1 + y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k$. Here,$n = 6$ and $y = (x

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$11$

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(C) The expansion is given by $(1 + (x - x^2))^6$. Using the binomial theorem,$(1 + y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k$. Here,$n = 6$ and $y = (x - x^2)$. $(1 + x - x^2)^6 = \sum_{k=0}^{6} \binom{6}{k} (x - x^2)^k = \sum_{k=0}^{6} \binom{6}{k} x^k (1 - x)^k$. To find the coefficient of $x^4$: For $k=2$: $\binom{6}{2} x^2 (1 - x)^2 = 15 x^2 (1 - 2x + x^2) = 15x^2 - 30x^3 + 15x^4$. Coefficient is $15$. For $k=3$: $\binom{6}{3} x^3 (1 - x)^3 = 20 x^3 (1 - 3x + 3x^2 - x^3) = 20x^3 - 60x^4 + 60x^5 - 20x^6$. Coefficient is $-60$. For $k=4$: $\binom{6}{4} x^4 (1 - x)^4 = 15 x^4 (1 - 4x + 6x^2 - 4x^3 + x^4) = 15x^4 - 60x^5 + 90x^6 - 60x^7 + 15x^8$. Coefficient is $15$. Total coefficient of $x^4 = 15 - 60 + 15 = -30$. To find the coefficient of $x^6$: For $k=3$: $\binom{6}{3} x^3 (1 - x)^3 = 20 x^3 (1 - 3x + 3x^2 - x^3) = 20x^3 - 60x^4 + 60x^5 - 20x^6$. Coefficient is $-20$. For $k=4$: $\binom{6}{4} x^4 (1 - x)^4 = 15 x^4 (1 - 4x + 6x^2 - 4x^3 + x^4) = 15x^4 - 60x^5 + 90x^6 - 60x^7 + 15x^8$. Coefficient is $90$. For $k=5$: $\binom{6}{5} x^5 (1 - x)^5 = 6 x^5 (1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5) = 6x^5 - 30x^6 + 60x^7 - 60x^8 + 30x^9 - 6x^{10}$. Coefficient is $-30$. For $k=6$: $\binom{6}{6} x^6 (1 - x)^6 = 1 x^6 (1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) = x^6 - 6x^7 + 15x^8 - 20x^9 + 15x^{10} - 6x^{11} + x^{12}$. Coefficient is $1$. Total coefficient of $x^6 = -20 + 90 - 30 + 1 = 41$. Sum of coefficients $= -30 + 41 = 11$.

Sum of the coefficients of $x^4$ and $x^6$ in the expansion of $(1+x-x^2)^6$ is

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