The coefficient of x^7 in the expansion of (1 | vedclass

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Question

(B) Given expression: $(1 - x - x^2 + x^3)^6 = ((1 - x)(1 - x^2))^6 = (1 - x)^6 (1 - x^2)^6 = (1 - x)^6 (1 - x)^6 (1 + x)^6 = (1 - x)^{12} (1 + x)^6$.

Vedclass · Accepted Answer

$-144$

Vedclass · Answer

(B) Given expression: $(1 - x - x^2 + x^3)^6 = ((1 - x)(1 - x^2))^6 = (1 - x)^6 (1 - x^2)^6 = (1 - x)^6 (1 - x)^6 (1 + x)^6 = (1 - x)^{12} (1 + x)^6$.We need the coefficient of $x^7$ in $(1 - x)^{12} (1 + x)^6$.Let $(1 - x)^{12} = \sum_{i=0}^{12} (-1)^i \binom{12}{i} x^i$ and $(1 + x)^6 = \sum_{j=0}^{6} \binom{6}{j} x^j$.The coefficient of $x^7$ is $\sum_{i+j=7} (-1)^i \binom{12}{i} \binom{6}{j}$.For $i=1, j=6: -\binom{12}{1} \binom{6}{6} = -12 	imes 1 = -12$.For $i=2, j=5: \binom{12}{2} \binom{6}{5} = 66 	imes 6 = 396$.For $i=3, j=4: -\binom{12}{3} \binom{6}{4} = -220 	imes 15 = -3300$.For $i=4, j=3: \binom{12}{4} \binom{6}{3} = 495 	imes 20 = 9900$.For $i=5, j=2: -\binom{12}{5} \binom{6}{2} = -792 	imes 15 = -11880$.For $i=6, j=1: \binom{12}{6} \binom{6}{1} = 924 	imes 6 = 5544$.For $i=7, j=0: -\binom{12}{7} \binom{6}{0} = -792 	imes 1 = -792$.Summing these: $-12 + 396 - 3300 + 9900 - 11880 + 5544 - 792 = -144$.

The coefficient of $x^7$ in the expansion of $(1 - x - x^2 + x^3)^6$ is

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