If (1+x+x^2+x^3)^5 = sum_{k=0}^{15} a_k x^k,t | vedclass

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(C) Given,$(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$. We can factor the expression as $(1+x)(1+x^2)^5 = (1+x)^5(1+x^2)^5$. Alternatively,$(1+x+x^2+x^

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$512$

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(C) Given,$(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$. We can factor the expression as $(1+x)(1+x^2)^5 = (1+x)^5(1+x^2)^5$. Alternatively,$(1+x+x^2+x^3)^5 = [(1+x)(1+x^2)]^5 = (1+x)^5(1+x^2)^5$. Let $f(x) = (1+x)^5(1+x^2)^5 = \sum_{k=0}^{15} a_k x^k$. We want to find $\sum_{k=0}^7 a_{2k} = a_0 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} + a_{14}$. Note that $f(1) = \sum_{k=0}^{15} a_k = (1+1)^5(1+1^2)^5 = 2^5 	imes 2^5 = 2^{10} = 1024$. And $f(-1) = \sum_{k=0}^{15} a_k (-1)^k = (1-1)^5(1+(-1)^2)^5 = 0^5 	imes 2^5 = 0$. Thus,$f(1) + f(-1) = 2(a_0 + a_2 + a_4 + \dots + a_{14}) = 1024 + 0 = 1024$. Therefore,$a_0 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} + a_{14} = \frac{1024}{2} = 512$.

If $(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$,then $\sum_{k=0}^7 a_{2k}$ is equal to

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