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$\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=$
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Verified Answer
The correct answer is:
$\frac{2^{n+1}-1}{n+1}$
Proceeding as above and putting $n+1=N$.
So given term can be written as
$\begin{aligned}
& \frac{1}{N}\left\{{ }^N C_1+{ }^N C_2+{ }^N C_3+\ldots .\right\} \\
= & \frac{1}{N}\left\{2^N-1\right\}=\frac{1}{n+1}\left(2^{n+1}-1\right) \quad(N=n+1)\end{aligned}$
So given term can be written as
$\begin{aligned}
& \frac{1}{N}\left\{{ }^N C_1+{ }^N C_2+{ }^N C_3+\ldots .\right\} \\
= & \frac{1}{N}\left\{2^N-1\right\}=\frac{1}{n+1}\left(2^{n+1}-1\right) \quad(N=n+1)\end{aligned}$
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