Let f n = 0 2 ( k = 1 n sin k - 1 x ) ( k = 1 n ( 2 k - 1 ) sin k - 1 x ) cos x ? d x , n . Then f 21 - f 20 is equal to ________. [2023]

Question

Let $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, n \in ℕ .$ Then $f_{21} - f_{20}$ is equal to ________. [2023]

Smart Achievers · Accepted Answer

(41)

Let $f_{n} = \int_{0}^{π / 2} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, x \in ℕ$

$f_{n} (x) = \int_{0}^{π / 2} (1 + \sin x + \sin^{2} x + \dots + \sin^{n - 1} x) (1 + 3 \sin x + 5 \sin^{2} x + \dots + (2 n - 1) \sin^{n - 1} x) \cdot \cos x d x$

Multiply and divide by $\sqrt{\sin x}$ , we get:

$\int_{0}^{π / 2} [{(\sin x)}^{1 / 2} + {(\sin x)}^{3 / 2} + {(\sin x)}^{5 / 2} + \dots + {(\sin x)}^{\frac{2 n - 1}{2}}] (1 + 3 \sin x + 5 \sin^{2} x + \dots + (2 n - 1) \sin^{n - 1} x) \frac{\cos x}{\sqrt{\sin x}} d x$

Put ${(\sin x)}^{1 / 2} + {(\sin x)}^{3 / 2} + {(\sin x)}^{5 / 2} + \dots + {(\sin x)}^{n - \frac{1}{2}} = t$

$\Rightarrow \frac{1}{2} \frac{(1 + 3 \sin x + 5 \sin^{2} x + \dots + (2 n - 1) \sin^{n - 1} x)}{\sqrt{\sin x}} \cos x d x = d t$

$\Rightarrow f_{n} = 2 \int_{0}^{n} t d t \Rightarrow f_{n} = n^{2}$

$∴$ $f_{21} - f_{20} = {(21)}^{2} - {(20)}^{2} = 441 - 400 = 41$

Solution

Q.
Let $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, n \in ℕ .$ Then $f_{21} - f_{20}$ is equal to ________. [2023]

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Solution

Q. Let fn=∫0π2(∑k=1nsink-1x)(∑k=1n(2k-1)sink-1x)cosx dx, n∈ℕ. Then f21-f20 is equal to ________. [2023]

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Q.
Let $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x, n \in ℕ .$ Then $f_{21} - f_{20}$ is equal to ________. [2023]