मान लीजिए f_n = int_0^{frac{pi}{2}} left(sum_ | vedclass

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(B) मान लीजिए $u = \sin x$,तब $du = \cos x \, dx$। जब $x=0, u=0$ और जब $x=\frac{\pi}{2}, u=1$।$f_n = \int_0^1 \left(\sum_{k=1}^n u^{k-1}\right) \left(

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(B) मान लीजिए $u = \sin x$,तब $du = \cos x \, dx$। जब $x=0, u=0$ और जब $x=\frac{\pi}{2}, u=1$।$f_n = \int_0^1 \left(\sum_{k=1}^n u^{k-1}\right) \left(\sum_{k=1}^n (2k-1) u^{k-1}\right) du$।मान लीजिए $S_1 = \sum_{k=1}^n u^{k-1} = 1 + u + u^2 + \dots + u^{n-1} = \frac{1-u^n}{1-u}$।मान लीजिए $S_2 = \sum_{k=1}^n (2k-1) u^{k-1} = \frac{d}{du} \sum_{k=1}^n u^{2k-1} = \frac{d}{du} (u + u^3 + \dots + u^{2n-1}) = \frac{d}{du} \left( u \frac{1-u^{2n}}{1-u^2} \right)$।समाकलन करने पर,हमें प्राप्त होता है $f_n = n^2$।अतः,$f_{21} - f_{20} = 21^2 - 20^2 = (21-20)(21+20) = 41$।

मान लीजिए $f_n = \int_0^{\frac{\pi}{2}} \left(\sum_{k=1}^n \sin^{k-1} x\right) \left(\sum_{k=1}^n (2k-1) \sin^{k-1} x\right) \cos x \, dx$,जहाँ $n \in N$ है। तो $f_{21} - f_{20}$ का मान $...........$ है।

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