Question

The value of $\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{r+{}^3{C}_r}$ is

• A. $\frac{3!}{n+3}$
• B. $\frac{3!}{2(n+3)}$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $\frac{3!}{2(n+3)}$

Let us assume that ${\sum }_{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+3}{C}_r}$ is $\lambda$.
Hence, $\lambda$ $=\frac{{}^1{C}_0}{{}^3{C}_0}-\frac{{}^n{C}_1}{{}^4{C}_1}+\frac{{}^n{C}_2}{{}^5{C}_2}...$
$=1-\frac{n}{4}+\frac{n(n-1)}{20}...$
Now for $n=1$
$\lambda$ $=1-\frac{1}{4}$ $=\frac{3}{4}$
Therefore, we can see that only option $B$ yields the above answer for $n=1$.
Hence answer is Option B