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Let $S=\frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}$ $+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n} .$ Then, $S$ equals
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Verified Answer
The correct answer is:
$\frac{3^{n+1}-1}{n+1}$
We know that $(1+x)^{n}={ }^{n} C_{0}+x{ }^{-} C_{1}+x^{2}{ }^{n} C_{2}+\ldots+x^{n}{ }^{n} C_{n}$
On integrating both sides from 0 to 2 , we get
$\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{2}$
$=\left[x^{n} C_{0}+\frac{x^{2}}{2}^{n} C_{1}+\frac{x^{3}}{3}^{n} C_{2}+\ldots+\frac{x^{n+1}}{n+1}^{n} C_{n}\right]_{0}^{2}$
$\Rightarrow \frac{(3)^{n+1}}{n+1}-\frac{1}{n+1}=2{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}$
$\quad+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}-0$
$\Rightarrow \frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}$
$\quad=\frac{3^{n+1}-1}{n+1}$
On integrating both sides from 0 to 2 , we get
$\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{2}$
$=\left[x^{n} C_{0}+\frac{x^{2}}{2}^{n} C_{1}+\frac{x^{3}}{3}^{n} C_{2}+\ldots+\frac{x^{n+1}}{n+1}^{n} C_{n}\right]_{0}^{2}$
$\Rightarrow \frac{(3)^{n+1}}{n+1}-\frac{1}{n+1}=2{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}$
$\quad+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}-0$
$\Rightarrow \frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}$
$\quad=\frac{3^{n+1}-1}{n+1}$
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