(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} $ ……$(i)$

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} – ….} \right]$

$ = \frac{{{C_0}}}{2} – \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} – ….$

$\therefore \,\,\,\,\frac{{{C_0}}}{2} – \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} – ….$to $(n + 1)$ terms

$ = \frac{1}{{n + 1}} – \frac{1}{{n + 2}} = \frac{1}{{(n + 1)(n + 2)}}$.