Question

The value of $\frac{C_0}{1.3}$ - $\frac{C_1}{2.3}$+ $\frac{C_2}{3.3}$- $\frac{C_3}{4.3}$+ ..................+$(-1{)}^n$ $\frac{C_n}{(n+1)}$$\cdot 3$

• A. $\frac{3}{n+1}$
• B. $\frac{n+3}{3}$
• C. $\frac{1}{3(n+1)}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{3(n+1)}$

$\frac{C_0}{1.3}-\frac{C_1}{2.3}+\frac{C_2}{3.3}-\frac{C_3}{4.3}+....+{(-1)}^n\frac{C_n}{(n+1)}.3$

LHS $=1-\frac{n}{2}+\frac{n(n-1)}{3\times 2!}.......+(-1{)}^n\frac{1}{n+1}$

$=\frac{1}{3(n+1)}\left[\right(n+1)-\frac{(n-1)n}{2}-\frac{(n-1)(n+1)n}{3}......(-1{)}^{n}1]$

$=\frac{1}{3(n+1)}[{}^{n+1}{C}_1-{}^{n+1}{C}_2+{}^{n+1}{C}_3-{}^{n+1}{C}_4+......(-1{)}^n.{}^{n+1}{C}_{n+1}]$

$=\frac{1}{3(n+1)}[1-({}^{n+1}{C}_0-{}^{n+1}{C}_1+{}^{n+1}{C}_2-......(-1{)}^n.{}^{n+1}{C}_{n+1}\left)\right]$

$=\frac{1}{3(n+1)}[1-{(1-1)}^{n+1}]$

$=\frac{1}{3(n+1)}[1-0]$

$=\frac{1}{3(n+1)}$

Hence, the answer is $\frac{1}{3(n+1)}.$