Discussion Forum

If n is a positive integer greater than 1, then  $3(^nC_{0})-8(^nC_{1})+13(^nC_{2})-18(^nC_{3})+...$    $upto (n+1)$ terms =

Answer:

Explanation:

 The general  term of the  given series  is 

$T_{r}=(-1)^{r}(3+5r)^{n}C_{r},r=0,1,2,.....,n$

 $\therefore$    $S_{n}=\sum_{r=0}^{n}(-1)^{r}(3+5r)^{n} C_{r}$

 $=3\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}+5\sum_{r=0}^{n}(-1)^{r}.{r^{n}}C_{r}$

    $=3\left[\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}\right]+5\left[\sum_{r=1}^{n}(-1)^{r}.{r^{}\frac{n}{r}}.{^{n-1}}C_{r-1}\right]$

$=3\left[\sum_{r=0}^{n}(-1)^{r}.{^{n}}C_{r}\right]+5n\left[\sum_{r=1}^{n}(-1)^{r}.{^{n-1}}C_{r-1}\right]$

 =$3(0)+5n(0)=0+0=0$