The sum to (n + 1) terms of the series frac{C | vedclass

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(C) We know that $(1 - x)^n = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + (-1)^n C_n x^n$.Multiplying by $x$,we get $x(1 - x)^n = C_0x - C_1x^2 + C_2x^3 -

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$\frac{1}{(n + 1)(n + 2)}$

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(C) We know that $(1 - x)^n = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + (-1)^n C_n x^n$.Multiplying by $x$,we get $x(1 - x)^n = C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots + (-1)^n C_n x^{n+1}$.Integrating both sides from $0$ to $1$:$\int_0^1 x(1 - x)^n dx = \int_0^1 (C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots) dx$.On the $R$.$H$.$S$.,we get $\left[ \frac{C_0x^2}{2} - \frac{C_1x^3}{3} + \frac{C_2x^4}{4} - \dots \right]_0^1 = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \dots$.On the $L$.$H$.$S$.,let $1 - x = t$,so $dx = -dt$. When $x=0, t=1$ and when $x=1, t=0$.$\int_1^0 (1 - t)t^n (-dt) = \int_0^1 (t^n - t^{n+1}) dt = \left[ \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2} \right]_0^1 = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}$.Thus,the sum is $\frac{1}{(n+1)(n+2)}$.

The sum to $(n + 1)$ terms of the series $\frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \dots$ is

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