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Question: \(\left( 2.^{10}C_{0} \right)\) + \(\left( \frac{2^{2}}{2}.^{10}C_{1} \right)\) + \(\left( \frac{2^{...

(2.10C0)\left( 2.^{10}C_{0} \right) + (222.10C1)\left( \frac{2^{2}}{2}.^{10}C_{1} \right) + (233.10C2)\left( \frac{2^{3}}{3}.^{10}C_{2} \right) +.........+(21111.10C10)\left( \frac{2^{11}}{11}.^{10}C_{10} \right)=

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

∫02(1+x)10dx=∫02(C0\int_{0}^{2}{(1 + x)^{10}dx =}\int_{0}^{2}{(C_{0}}+ C1x+C2x2+.....+C10 x10)dx

= (1+2)11–111\frac{(1 + 2)^{11}–1}{11} = C0.(2)1\frac{C_{0}.(2)}{1} + C1.222\frac{C_{1}.2^{2}}{2} + C2.233\frac{C_{2}.2^{3}}{3}+ ......... +C10.21111\frac{C_{10}.2^{11}}{11}

= 311–111\frac{3^{11}–1}{11}