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+1 vote
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in Mathematics by (47.3k points)
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The value of k ∈ N for which the integral \(I_{n}=\int_{0}^{1}\left(1-x^{k}\right)^{n} d x, n \in \mathbb{N}, \) satisfies \(147 I_{20}=148 I_{21}\) is :

(1) 10 

(2) 8 

(3) 14 

(4) 7

1 Answer

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Best answer

Correct option is (4) 7  

\(\mathrm{I}_{\mathrm{n}}=\int_{0}^{1}\left(1-\mathrm{x}^{\mathrm{k}}\right)^{\mathrm{n}} \cdot 1 \mathrm{dx}\)

\(I_{n}=\left(1-x^{k}\right)^{n} \cdot x-n k \int_{0}^{1}\left(1-x^{k}\right)^{n-1} \cdot x^{k-1} \cdot d x\)

\(\mathrm{I}_{\mathrm{n}}=\mathrm{nk} \int_{0}^{1}\left[\left(1-\mathrm{x}^{\mathrm{k}}\right)^{\mathrm{n}}-\left(1-\mathrm{x}^{\mathrm{k}}\right)^{\mathrm{n}-1}\right] \mathrm{dx}\)

\(\mathrm{I}_{\mathrm{n}}=\mathrm{nkI}_{\mathrm{n}}-\mathrm{nkI}_{\mathrm{n}}\)

\(\frac{I_{n}}{I_{n-1}}=\frac{n k}{n k+1}\)

\(\frac{\mathrm{I}_{21}}{\mathrm{I}_{20}}=\frac{21 \mathrm{k}}{1+21 \mathrm{k}}\)

\(=\frac{147}{148} \Rightarrow \mathrm{k}=7\)

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