Let f_n = int_0^{frac{pi}{2}} left(sum_{k=1}^ | vedclass

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(B) Let $u = \sin x$,then $du = \cos x \, dx$. When $x=0, u=0$ and when $x=\frac{\pi}{2}, u=1$.$f_n = \int_0^1 \left(\sum_{k=1}^n u^{k-1}\right) \left

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$41$

Vedclass · Answer

(B) Let $u = \sin x$,then $du = \cos x \, dx$. When $x=0, u=0$ and when $x=\frac{\pi}{2}, u=1$.$f_n = \int_0^1 \left(\sum_{k=1}^n u^{k-1}\right) \left(\sum_{k=1}^n (2k-1) u^{k-1}\right) du$.Let $S_1 = \sum_{k=1}^n u^{k-1} = 1 + u + u^2 + \dots + u^{n-1} = \frac{1-u^n}{1-u}$.Let $S_2 = \sum_{k=1}^n (2k-1) u^{k-1} = \frac{d}{du} \sum_{k=1}^n u^{2k-1} = \frac{d}{du} (u + u^3 + \dots + u^{2n-1}) = \frac{d}{du} \left( u \frac{1-u^{2n}}{1-u^2} \right)$.Alternatively,note that $f_n = \int_0^1 \left(\sum_{i=0}^{n-1} u^i\right) \left(\sum_{j=1}^n (2j-1) u^{j-1}\right) du$.By evaluating the integral,we find $f_n = n^2$.Proof: $f_n - f_{n-1} = \int_0^1 \left( (\sum_{k=1}^n u^{k-1})(\sum_{k=1}^n (2k-1) u^{k-1}) - (\sum_{k=1}^{n-1} u^{k-1})(\sum_{k=1}^{n-1} (2k-1) u^{k-1}) \right) du = 2n-1$.Thus,$f_n = \sum_{k=1}^n (2k-1) = n^2$.Therefore,$f_{21} - f_{20} = 21^2 - 20^2 = (21-20)(21+20) = 41$.

Let $f_n = \int_0^{\frac{\pi}{2}} \left(\sum_{k=1}^n \sin^{k-1} x\right) \left(\sum_{k=1}^n (2k-1) \sin^{k-1} x\right) \cos x \, dx$,where $n \in N$. Then $f_{21} - f_{20}$ is equal to $...........$.

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