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$^nC_0+2^nC_1+3^nC_2+\cdots+(n+1)^nC_n$ equals

1. $2^n+n2^{n-1}$
2. $2^n-n2^{n-1}$
3. $2^n$
4. none of these

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+1 vote

Let us take this binomial expansion:

$\left ( 1 + x \right )^{^{n}} = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2}x^{2} + \binom{n}{3}x^{3} + \cdot + \cdot + \binom{n}{n}x^{n}$

Multiplying it with x gives:

$x\left ( 1 + x \right )^{^{n}} = \binom{n}{0}x + \binom{n}{1} x^{2} + \binom{n}{2}x^{3} + \binom{n}{3}x^{4} + \cdot + \cdot + \binom{n}{n}x^{n+1}$

Now differentiate both sides w.r.t x:

$x\left [ n\left ( 1 + x \right )^{n-1} \right ]+ \left ( 1 + x \right )^{^{n}} = \binom{n}{0} + 2\binom{n}{1} x^{1} + 3\binom{n}{2}x^{2} + 4\binom{n}{3}x^{3} + \cdot + \cdot + \left ( n+1 \right )\binom{n}{n}x^{n}$

Now, substitute x = 1:

$\left [ n\left ( 2 \right )^{n-1} \right ]+ \left ( 2 \right )^{^{n}} = \binom{n}{0} + 2\binom{n}{1} + 3\binom{n}{2} + 4\binom{n}{3} + \cdot + \cdot + \left ( n+1 \right )\binom{n}{n}$

Thus, the answer is option (A).

edited
+1

There are two mistakes above these lines.

+1
Thank you for pointing out the mistakes. I have corrected them.
+1 vote
Let $I=\binom{n}{0}+2*\binom{n}{1}+3*\binom{n}{2}+...+(n+1)\binom{n}{n}$$----->1 Rewrite this I as I=(n+1)\binom{n}{n}+n\binom{n}{n-1}+(n-1)\binom{n}{n-2}...+\binom{n}{0} I=(n+1)\binom{n}{0}+n\binom{n}{1}+(n-1)\binom{n}{2}...+\binom{n}{n}$$------>2$    $\because \binom{n}{k}=\binom{n}{n-k}$

adding corresponding terms of equations 1 and 2,

$2I=(n+2)\binom{n}{0}+(n+2)\binom{n}{1}+(n+2)\binom{n}{2}$...$+(n+2)\binom{n}{n}$

$2I=(n+2)(\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...\binom{n}{n})$

$2I=(n+2)(1+1)^{n}$

$2I=(n+2)2^{n}$

$I=(n+2)2^{n-1}$

$I=n2^{n-1}+2^{n}$

Hence option A is the answer
by Loyal