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Mathematics Question on Integrals of Some Particular Functions

Let S=21nC0+222nC1+233nC2+......+2n+1n+1nCnS=\frac{2}{1} ^{n}C_{0}+\frac{2^{2}}{2} ^{n}C_{1}+\frac{2^{3}}{3} ^{n}C_{2}+ ...... +\frac{2^{n+1}}{n+1} ^{n}C_{n}. Then SS equals

A

2n+1−1n+1\frac{2^{n+1}-1}{n+1}

B

3n+1−1n+1\frac{3^{n+1}-1}{n+1}

C

3n−1n\frac{3^n-1}{n}

D

2n−1n\frac{2^{n}-1}{n}

Answer

3n+1−1n+1\frac{3^{n+1}-1}{n+1}

Explanation

Solution

We know that
(1+x)n=nC0+xnC1+x2nC2+…+xnnCn(1+x)^{n}={ }^{n} C_{0}+x{ }^{n} C_{1}+x^{2}{ }^{n} C_{2}+\ldots+x^{n}{ }^{n} C_{n}
On integrating both sides from 0 to 2 , we get
[(1+x)n+1n+1]02\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{2}
=[xnC0+x22nC1+x33nC2+…+xn+1n+1nCn]02=\left[x^{n} C_{0}+\frac{x^{2}}{2}{ }^{n} C_{1}+\frac{x^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{x^{n+1}}{n+1}{ }^{n} C_{n}\right]_{0}^{2}
⇒(3)n+1n+1−1n+1=2nC0+222nC1+233nC2+…+2n+1n+1nCn−0\Rightarrow \frac{(3)^{n+1}}{n+1}-\frac{1}{n+1}=2{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}-0
⇒21nC0+222nC1+233nC2+…+2n+1n+1nCn\Rightarrow \frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}
=3n+1−1n+1=\frac{3^{n+1}-1}{n+1}