2{C_0} + frac{2^2}{2}{C_1} + frac{2^3}{3}{C_2 | vedclass

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(A) We know the binomial expansion: $(1 + x)^{10} = \sum_{r=0}^{10} {C_r} x^r = {C_0} + {C_1}x + {C_2}x^2 + \dots + {C_{10}}x^{10}$.Integrating both s

Vedclass · Accepted Answer

$\frac{3^{11} - 1}{11}$

Vedclass · Answer

(A) We know the binomial expansion: $(1 + x)^{10} = \sum_{r=0}^{10} {C_r} x^r = {C_0} + {C_1}x + {C_2}x^2 + \dots + {C_{10}}x^{10}$.Integrating both sides with respect to $x$ from $0$ to $2$:$\int_{0}^{2} (1 + x)^{10} dx = \int_{0}^{2} ({C_0} + {C_1}x + {C_2}x^2 + \dots + {C_{10}}x^{10}) dx$.Evaluating the integral on the left side:$\left[ \frac{(1 + x)^{11}}{11} \right]_{0}^{2} = \frac{(1 + 2)^{11}}{11} - \frac{(1 + 0)^{11}}{11} = \frac{3^{11} - 1}{11}$.Evaluating the integral on the right side:$\left[ {C_0}x + {C_1}\frac{x^2}{2} + {C_2}\frac{x^3}{3} + \dots + {C_{10}}\frac{x^{11}}{11} \right]_{0}^{2} = 2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}}$.Thus,the sum is equal to $\frac{3^{11} - 1}{11}$.

$2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}} = \dots$

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