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3 Answers

3 votes
3 votes

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 by
2 votes
2 votes
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
0 votes
0 votes
The answer must be none of these as none of the option satisfies the condition for the given value of N , if we put N==3 we get 170 , but when we try to put the n==3 and try to match the options for the given value of n=3 we get no matches hence answer must be D none of these

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