The coefficient of x^{4} in the expansion of | vedclass

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(C) We have $(1+x+x^{2}+x^{3})^{6} = ((1+x)(1+x^{2}))^{6} = (1+x)^{6}(1+x^{2})^{6}$.Using the binomial expansion,$(1+x)^{6} = \sum_{r=0}^{6} {}^{6}C_{

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) We have $(1+x+x^{2}+x^{3})^{6} = ((1+x)(1+x^{2}))^{6} = (1+x)^{6}(1+x^{2})^{6}$.Using the binomial expansion,$(1+x)^{6} = \sum_{r=0}^{6} {}^{6}C_{r} x^{r}$ and $(1+x^{2})^{6} = \sum_{t=0}^{6} {}^{6}C_{t} x^{2t}$.The product is $\sum_{r=0}^{6} \sum_{t=0}^{6} {}^{6}C_{r} {}^{6}C_{t} x^{r+2t}$.To find the coefficient of $x^{4}$,we set $r+2t = 4$. The possible non-negative integer solutions $(r, t)$ are:$r$$t$$0$$2$$2$$1$$4$$0$The coefficient is ${}^{6}C_{0} 	imes {}^{6}C_{2} + {}^{6}C_{2} 	imes {}^{6}C_{1} + {}^{6}C_{4} 	imes {}^{6}C_{0}$.$= (1 	imes 15) + (15 	imes 6) + (15 	imes 1) = 15 + 90 + 15 = 120$.

The coefficient of $x^{4}$ in the expansion of $(1+x+x^{2}+x^{3})^{6}$ in powers of $x$ is:

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