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  4. Let S=(2/1) nC0+(22/2) nC1+(23/3) nC2+ ...... +(2n+1/n+1) nCn. Then S equals

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Q. Let $S=\frac{2}{1} ^{n}C_{0}+\frac{2^{2}}{2} ^{n}C_{1}+\frac{2^{3}}{3} ^{n}C_{2}+ ...... +\frac{2^{n+1}}{n+1} ^{n}C_{n}$. Then $S$ equals

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Solution:

We know that
$(1+x)^{n}={ }^{n} C_{0}+x{ }^{n} C_{1}+x^{2}{ }^{n} C_{2}+\ldots+x^{n}{ }^{n} C_{n}$
On integrating both sides from 0 to 2 , we get
$\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{2}$
$=\left[x^{n} C_{0}+\frac{x^{2}}{2}{ }^{n} C_{1}+\frac{x^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{x^{n+1}}{n+1}{ }^{n} C_{n}\right]_{0}^{2}$
$\Rightarrow \frac{(3)^{n+1}}{n+1}-\frac{1}{n+1}=2{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}-0$
$\Rightarrow \frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}$
$=\frac{3^{n+1}-1}{n+1}$