Dear student 2C0+222C1+233C2+ +21111C10Consider,(1+x)10=C0+C1x+C2x2+ +C10x10Integrating both sides from 0 to 2, we get(1+x)111102=C0x+C1x22+C2x33+ +C10x111102⇒311-111=2C0+22C12+23C23+ +211C1011 Regards

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2 C_{0} + \frac{2^{2}}{2} C_{1} + \frac{2^{3}}{3} C_{2} + . . . . + \frac{2^{11}}{11} C_{10} Consider, (1 + x)^{10} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{10} x^{10} Integrating both sides from 0 to 2, we get {[\frac{(1 + x)^{11}}{11}]}_{0}^{2} = {[C_{0} x + \frac{C_{1} x^{2}}{2} + \frac{C_{2} x^{3}}{3} + . . . + \frac{C_{10} x^{11}}{11}]}_{0}^{2} \Rightarrow \frac{3^{11} - 1}{11} = 2 C_{0} + \frac{2^{2} C_{1}}{2} + \frac{2^{3} C_{2}}{3} + . . . + \frac{2^{11} C_{10}}{11}

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