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Step-by-Step Detailed Solution
Step 1: Express the given integral
We have
f_{n} = \int_{0}^{\frac{\pi}{2}} \Bigl(\sum_{k=1}^{n} \sin^{k-1} x\Bigr)\Bigl(\sum_{k=1}^{n}(2k-1)\,\sin^{k-1} x\Bigr)\cos x \, dx.
We want to find f_{21} - f_{20} .
Step 2: Use the substitution t = \sin x
Let t = \sin x . Then dt = \cos x\,dx .
As x ranges from 0 to \frac{\pi}{2} , t ranges from 0 to 1 .
Hence,
f_{n} = \int_{0}^{1} \Bigl(\sum_{k=1}^{n} t^{k-1}\Bigr)\Bigl(\sum_{k=1}^{n} (2k-1)\,t^{k-1}\Bigr)\,dt.
Step 3: Identify the terms for f_{21} and f_{20}
For n=21 :
\displaystyle \sum_{k=1}^{21} t^{k-1}
= 1 + t + t^2 + \cdots + t^{20},
\displaystyle \sum_{k=1}^{21} (2k-1)\,t^{k-1}
= 1 + 3t + 5t^{2} + \cdots + 41t^{20}.
For n=20 :
\displaystyle \sum_{k=1}^{20} t^{k-1}
= 1 + t + t^2 + \cdots + t^{19},
\displaystyle \sum_{k=1}^{20} (2k-1)\,t^{k-1}
= 1 + 3t + 5t^{2} + \cdots + 39t^{19}.
Step 4: Focus on the difference f_{21} - f_{20}
The integrals for f_{21} and f_{20} differ only in the highest power of t present in the sums:
• In f_{21} , there is an extra t^{20} term in \sum t^{k-1} and a corresponding extra 41\,t^{20} term in \sum (2k-1)\,t^{k-1} .
• These extra terms do not appear in f_{20} .
Hence, when you subtract f_{20} from f_{21} , all lower powers (which are identical in both expressions) cancel out, and only the contribution from those extra terms at t^{20} remains.
Step 5: Evaluate the extra term
The extra product of terms is:
\bigl(t^{20}\bigr) \times \bigl(41\,t^{20}\bigr) = 41\,t^{40}.
So, the extra contribution to the integral in f_{21} is
\int_{0}^{1} 41\,t^{40}\,dt = 41 \int_{0}^{1} t^{40}\,dt.
Since
\int_{0}^{1} t^{40}\,dt = \frac{1}{41},
we get
41 \times \frac{1}{41} = 1.
But we must check for any other cross-terms involving t^{20} in one sum and all terms in the other sum. Let’s inspect carefully:
• From \sum_{k=1}^{21} t^{k-1} , the new term is t^{20} .
• From \sum_{k=1}^{21} (2k-1)\,t^{k-1} , the new term is 41\,t^{20} .
When you multiply these sums out, you indeed get the product 41\,t^{40} as the new highest power contribution. However, there could also be “cross-terms” where t^{20} from one sum multiplies earlier powers from the other. To be precise:
1. t^{20} (from the first sum) times all terms (2k-1)\,t^{k-1} from the second sum, for k=1 to 21. This includes k=21 , giving 41\,t^{20} \times t^{20} = 41\,t^{40} , but also k
