Let f = ( k=1^ ^k x) ( k-1 ^k x) x dx N. Then f11 is equal to: — Solution | JEE Mathematics

NVAMediumJEE 2023Definite Integrals

Let $f = \left( \sum_{k=1}^{\infty} \sin^k x \right) \left( \sum_{k-1} \sin^k x \right) \cos x \, dx \in N$ . Then $f_{11}$ is equal to:

Answer

Correct answer:41

Standard Method

Given: the solution gives

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

Find: $f_{11}$

Let

t = \sin x

so that

\cos x \, dx = dt

Then

f_n = \int_0^1 \left( \sum_{k=1}^{n} t^{k-1} \right) \left( \sum_{k=1}^{n} (2k-1) t^{k-1} \right) dt

From the extracted solution, the difference is written as

f_{n+1} - f_n = \int_0^1 \left( 1 + t + t^2 + \cdots + t^n \right) \left( 1 + 3t + 5t^2 + \cdots + (2n+1)t^n \right) dt

Putting $n = 20$ , it gives

f_{21} - f_{20} = \int_0^1 \left( 1 + 3t + 5t^2 + \cdots + 41t^{20} \right) dt

Therefore,

f_{21} - f_{20} = \left( \frac{1}{21} + \frac{3}{22} + \frac{5}{23} + \cdots + \frac{41}{40} \right)

The extracted solution concludes that this sum simplifies to $40 + 1 = 41$ .

Therefore, according to the solution, the required value is $41$ .

Using the answer key instead of the worked solution. Here the source answer says $50$ , but the solution concludes $41$ . When working is available, the solution is the primary source.
Missing the substitution $t = \sin x$ and forgetting that $\cos x \, dx = dt$ . That changes the integral limits from $x : 0 \text{ to } \frac{\pi}{2}$ into $t : 0 \text{ to } 1$ .
Confusing $f_{n+1}-f_n$ with $f_n$ . The extracted working is written for a difference expression, so directly reading it as $f_{11}$ without tracking the index gives the wrong quantity.

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