Question

The value of $\sum _{r=0}^{50}(-1{)}^r\frac{{}^{50}{C}_r}{r+2}$ is equal to

• A. $\frac{1}{50\times 51}$
• B. $\frac{1}{52\times 50}$
• C. $\frac{1}{52\times 51}$
• D. $\frac{1}{50\times 53}$

Answer 1

The correct option is C $\frac{1}{52\times 51}$

$T_{r+1}=(-1{)}^r\frac{{}^{50}{C}_r}{r+2}$
$=(-1{)}^r(r+1)\frac{{}^{50}{C}_r}{(r+1)(r+2)}$

$=(-1{)}^r(r+1)\frac{{}^{52}{C}_{r+2}}{51\times 52}$

$=(-1{)}^r\frac{\left[\right(r+2)-1]{}^{52}{C}_{r+2}}{51\times 52}$

$=(-1{)}^r\frac{[52\times {}^{51}{C}_{r+1}-{}^{52}{C}_{r+2}]}{51\times 52}$

$=\frac{[-52\times {}^{51}{C}_{r+1}(-1{)}^{r+1}-{}^{52}{C}_{r+2}(-1{)}^{r+2}]}{51\times 52}$

$\therefore \sum _{r=0}^{50}(-1{)}^r\frac{{}^{50}{C}_r}{r+2}=-52\frac{(1-1{)}^{51}-{}^{51}{C}_0}{51\times 52}-\frac{(1-1{)}^{52}-{}^{52}{C}_0+{}^{52}{C}_1}{51\times 52}$

$=\frac{1}{51}-\frac{1}{52}$
$=\frac{1}{51\times 52}$