श्रेणी frac{{{C_0}}}{2} - frac{{{C_1}}}{3} + | vedclass

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Question

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

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Vedclass · Answer

${(1 - x)^n} = {C_0} - {C_1}x + {C_2}{x^2} - {C_3}{x^3} + .....$

 $x\,\,{(1 - x)^n} = {C_0}x - {C_1}{x^2} + {C_2}{x^3} - {C_3}{x^4} + .....$

  $\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;...$(i)$

$LHS$ का समाकलन

$ = \int\limits_1^0 {(1 - t){t^n}( - dt),} $ $1 - x = t$ रखने पर

$ = \int\limits_0^1 {({t^n} - {t^{n + 1}})} \,dt = \frac{1}{{n + 1}} - \frac{1}{{n + 2}}$

$(i)$ के $RHS$ का समाकलन

$ = \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} - ....$$(n + 1)$ पदों तक

$ = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} = \frac{1}{{(n + 1)(n + 2)}}$.

श्रेणी $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + $.....के $(n + 1)$ पदों का योग है

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