Question

If $n$ is a positive integer greater than 1, then
$3\left({}^n{C}_0\right)-8\left({}^n{C}_1\right)+13\left({}^n{C}_2\right)-18\left({}^n{C}_3\right)+\dots$ upto $(n+1)$ terms $=$

• A. -5
• B. $\frac{2^{n+1}-1}{n}$
• C. $\frac{2^n-1}{2}$
• D. 0

Answer 1

Answer: (D)

The general term of the given series is
$T_r=(-1{)}^r(3+5r{)}^n{C}_{r^{\prime }}r=0,1,2,\dots ,n$
$\therefore S_n=\sum _{r=0}^n(-1{)}^r(3+5r{)}^n{C}_r$
$=3\sum _{r=0}^n(-1{)}^r\cdot {}^n{C}_r+5\sum _{r=0}^n(-1{)}^{r}r{}^n{C}_r$
$=3\left[\sum _{r=0}^n(-1{)}^r\cdot {}^n{C}_r\right]+5\left[\sum _{r=1}^n(-1{)}^r\cdot r\frac{n}{r}\cdot {}^{n-1}{C}_{r-1}\right]$
$=3\left[\sum _{r=0}^n(-1{)}^{rn}{C}_r\right]+5n\left[\sum _{r=1}^n(-1{)}^r\cdot {}^{n-1}{C}_{r-1}\right]$
$=3(0)+5n(0)=0+0=0$