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Question

Question: <sup>n</sup>C<sub>0</sub> + 3.<sup>n</sup>C<sub>1</sub> + 5.<sup>n</sup>C<sub>2</sub> + ......... + ...

nC0 + 3.nC1 + 5.nC2 + ......... + (2n + 1)nCn =

A

2n

B

2n + n.2n –1

C

2n(n + 1)

D

None

Answer

2n(n + 1)

Explanation

Solution

S = nC0 + 3.nC1 + 5.nC2 + .......+ (2n + 1)nCn

+S=(2n+1).nC0+(2n–1).nC1+....+1.nCn2S=(2n+2)[nC0+nC1+...+nCn]\frac{+ S = (2n + 1).^{n}C_{0} + (2n–1).^{n}C_{1} + .... + 1.^{n}C_{n}}{2S = (2n + 2)\lbrack^{n}C_{0} +^{n}C_{1} + ... +^{n}C_{n}\rbrack}

S = (n + 1). (2)n