Let (1 + x + x^2)^{20}(2x + 1) = a_0 + a_1x^1 | vedclass

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(B) Given the expression $(1 + x + x^2)^{20}(2x + 1) = \sum_{k=0}^{41} a_k x^k$. Integrating both sides from $0$ to $1$ with respect to $x$: $\int_0^1

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$\frac{3^{21} - 1}{21}$

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(B) Given the expression $(1 + x + x^2)^{20}(2x + 1) = \sum_{k=0}^{41} a_k x^k$. Integrating both sides from $0$ to $1$ with respect to $x$: $\int_0^1 (1 + x + x^2)^{20}(2x + 1) dx = \int_0^1 (\sum_{k=0}^{41} a_k x^k) dx$. Let $u = 1 + x + x^2$,then $du = (1 + 2x) dx$. Note that the integrand is $(1 + x + x^2)^{20}(1 + 2x) dx$. So,$\int_0^1 u^{20} du = [\frac{u^{21}}{21}]_0^1$. At $x=0, u=1$. At $x=1, u=3$. Thus,$\int_1^3 u^{20} du = \frac{3^{21} - 1^{21}}{21} = \frac{3^{21} - 1}{21}$. The right side is $\sum_{k=0}^{41} \frac{a_k}{k+1} = \frac{a_0}{1} + \frac{a_1}{2} + ... + \frac{a_{41}}{42}$. Therefore,the sum is $\frac{3^{21} - 1}{21}$.

Let $(1 + x + x^2)^{20}(2x + 1) = a_0 + a_1x^1 + a_2x^2 + ... + a_{41}x^{41}$,then $\frac{a_0}{1} + \frac{a_1}{2} + .... + \frac{a_{41}}{42}$ is equal to

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