Question

Find the sum of ${}^n{C}_0$ - $\frac{{}^n{C}_1}{2}$ + $\frac{{}^n{C}_2}{3}$ ...............

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

When observe the terms, we find that each term is of the form $(-1{)}^r$ . $\frac{{}^n{C}_r}{(r+1)}$

Consider ${}^{n+1}{C}_{r+1}$ = $\frac{n+1}{r+1}$${}^n{C}_r$

⇒ $\frac{{}^n{C}_r}{r+1}$ = $\frac{1}{n+1}$ ${}^{n+1}{C}_{r+1}$

⇒${}^n{C}_0$ - $\frac{{}^n{C}_1}{2}$ + $\frac{{}^n{C}_2}{3}$ ....................... = $\sum _{r=0}^n(-1{)}^r$ $\frac{{}^n{C}_r}{r+1}$

= $\frac{1}{n+1}$ $\sum _{r=0}^n(-1{)}^r$ ${}^{n+1}{C}_{r+1}$

= $\frac{1}{n+1}$${(}^{n+1}{C}_1{-}^{n+1}{C}_2-..............)$

=$\frac{1}{n+1}$$({-}^{n+1}{C}_0{-}^{n+1}{C}_1{+}^{n+1}{C}_2.............{+}^{n+1}{C}_0)$

[Adding and substracting n +${}^1{C}_0$ ]

= $\frac{1}{n+1}(-{(}^{n+1}{\text{C}}_0{-}^{n+1}{\text{C}}_1{+}^{n+1}{\text{C}}_2....){+}^{n+1}{\text{C}}_0)$

= $\frac{1}{n+1}(0+{(}^{n+1}{\text{C}}_0))$

${[}^{n+1}{\text{C}}_0{-}^{n+1}{\text{C}}_1{+}^{n+1}{\text{C}}_2....isthesumofthecoefficientsof(1-x{)}^{n+1}.Itiszero]$

⇒ sum = $\frac{1}{n+1}$ ${}^{n+1}{C}_0$

= $\frac{1}{n+1}$