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Question: \[2C_{0} + \frac{2^{2}}{2}C_{1} + \frac{2^{3}}{3}C_{2} + ...... + \frac{2^{11}}{11}C_{10} =\]...

2C0+222C1+233C2+......+21111C10=2C_{0} + \frac{2^{2}}{2}C_{1} + \frac{2^{3}}{3}C_{2} + ...... + \frac{2^{11}}{11}C_{10} =

A

311−111\frac{3^{11} - 1}{11}

B

211−111\frac{2^{11} - 1}{11}

C

113−111\frac{11^{3} - 1}{11}

D

112−111\frac{11^{2} - 1}{11}

Answer

311−111\frac{3^{11} - 1}{11}

Explanation

Solution

It is clear that it is a expansion of

(1+x)10=C0+C1x+C2x2+.....C10x10(1 + x)^{10} = C_{0} + C_{1}x + C_{2}x^{2} + .....C_{10}x^{10}

Integrating w.r.t. x both sides between the limit 0 to 2.

[(1+x)1111]02=C0[x]02+C1[x22]02+C2[x33]02+......+C10[x1111]02\left\lbrack \frac{(1 + x)^{11}}{11} \right\rbrack_{0}^{2} = C_{0}\lbrack x\rbrack_{0}^{2} + C_{1}\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{2} + C_{2}\left\lbrack \frac{x^{3}}{3} \right\rbrack_{0}^{2} + ...... + C_{10}\left\lbrack \frac{x^{11}}{11} \right\rbrack_{0}^{2}

⇒ 311−111=2C0+222.C1+232.C2+......+21111C10\frac{3^{11} - 1}{11} = 2C_{0} + \frac{2^{2}}{2}.C_{1} + \frac{2^{3}}{2}.C_{2} + ...... + \frac{2^{11}}{11}C_{10}.