Question

If $n$ is a positive integer greater than 1, then
$3\left({ }^{n} C_{0}\right)-8\left({ }^{n} C_{1}\right)+13\left({ }^{n} C_{2}\right)-18\left({ }^{n} C_{3}\right)+\ldots$ 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
$T_{r}=(-1)^{r}(3+5 r)^{n} C_{r'} r=0,1,2, \ldots, n$
$\therefore S_{n}=\displaystyle \sum_{r=0}^{n}(-1)^{r}(3+5 r)^{n} C_{r}$
$=3 \displaystyle \sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n} C_{r}+5 \sum_{r=0}^{n}(-1)^{r} r{ }^{n} C_{r}$
$=3\left[\displaystyle \sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n} C_{r}\right]+5\left[\displaystyle \sum_{r=1}^{n}(-1)^{r} \cdot r \frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right]$
$=3\left[\displaystyle \sum_{r=0}^{n}(-1)^{r n} C_{r}\right]+5 n\left[\displaystyle \sum_{r=1}^{n}(-1)^{r} \cdot{ }^{n-1} C_{r-1}\right]$
$=3(0)+5 n(0)=0+0=0$