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Question

Question: \[C_{1} + 2C_{2} + 3C_{3} + .....^{n}C_{n} =\]...

C1+2C2+3C3+.....nCn=C_{1} + 2C_{2} + 3C_{3} + .....^{n}C_{n} =

A

2n2^{n}

B

n.2nn.2^{n}

C

n.2n−1n.2^{n - 1}

D

n.2n+1n.2^{n + 1}

Answer

n.2n−1n.2^{n - 1}

Explanation

Solution

We know that, (1+x)n=C0+C1x+C2x2+......+Cnxn(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ...... + C_{n}x^{n}......(i)

Differentiating both sides w.r.t. x, we get n(1+x)n−1n(1 + x)^{n - 1}

= 0 + C1+2.C2x+3C3x2+.......+nCnxn−1C_{1} + 2.C_{2}x + 3C_{3}x^{2} + ....... + nC_{n}x^{n - 1}

Putting x=1,x = 1, we get, n.2n−1=C1+2C2+3C3+......+nCnn.2^{n - 1} = C_{1} + 2C_{2} + 3C_{3} + ...... + nC_{n}.