The sum to (n + 1) terms of the following ser | vedclass

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Question

(d) ${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + .....$

==> $x\,\,{(1 - x)^n} = {C_0}x - {C_1}{x^2} + {C_2}{x^3} - {C_3}{x^4} + ...

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(d) ${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + .....$

==> $x\,\,{(1 - x)^n} = {C_0}x - {C_1}{x^2} + {C_2}{x^3} - {C_3}{x^4} + .....$ 

==> $\int\limits_0^1 x {(1 - x)^n}dx = \int\limits_0^1 {({C_0}x - {C_1}{x^2} + {C_2}{x^3}....)dx} $ &hellip;...$(i)$

The integral on the $LHS$

$ = \int\limits_1^0 {(1 - t){t^n}( - dt),} $by putting $1 - x = t$

$ = \int\limits_0^1 {({t^n} - {t^{n + 1}})} \,dt = \frac{1}{{n + 1}} - \frac{1}{{n + 2}}$

Whereas the integral on the $RHS$ of $(i)$

$ = \left[ {\frac{{{C_0}{x^2}}}{2} - \frac{{{C_1}{x^3}}}{3} + \frac{{{C_2}{x^4}}}{4} - ....} ight]$

$ = \frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - ....$

$	herefore \,\,\,\,\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - ....$to $(n + 1)$ terms

$ = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} = \frac{1}{{(n + 1)(n + 2)}}$.

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

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