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+2 votes

+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)**.

+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

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

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