Question

The sum to (n + 1) terms of the series $\frac{c_0}{2}-\frac{c_1}{3}+\frac{c_2}{4}-\frac{c_3}{5}+....is$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-c_3{x}^3+....nterms\therefore x(1-x{)}^n=C_{0}x-C_1{x}^2+c_2{x}^3-c_3{x}^4+.....\therefore {\int }_0^{1}x(1-x{)}^{n}dx={\int }_0^1(C_{0}x-C_1{x}^2+C_2{x}^3-c_3{x}^4+...)dx\Rightarrow -{\int }_0^1(1-t)t^{n}dt$ $={\left|(C_0\frac{x^2}{2}-C_1\frac{x^3}{3}+C_2\frac{x^4}{4}-C_3\frac{x^5}{5}+....)\right|}_0^1(wherer=1-x)-{\left|(\frac{r^{n+1}}{n+1}-\frac{r^{1+2}}{n+2})\right|}_1^0=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+.....\therefore \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+.....=\frac{1}{n+2}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}$