Binomial Theorem 2 Question 7
10.
If $C_{r}$ stands for ${ }^{n} C_{r}$, then the sum of the series
$ \frac{2 (\frac{n}{2}) ! (\frac{n}{2}) !}{n !}\left[C_{0}^{2}-2 C_{1}^{2}+3 C_{2}^{2}-\ldots+(-1)^{n}(n+1) C_{n}^{2}\right] $
where $n$ is an even positive integer, is
(a) $(-1)^{n / 2}(n+2)$
(b) $(-1)^{n}(n+1)$
(c) $(-1)^{n/2}(n+1)$
(d) None of these
Show Answer
Answer:
Correct Answer: 10. (a)
Solution:
- We have,
$ \begin{aligned} C_{0}^{2}-2 & C_{1}^{2}+3 C_{2}^{2}-4 C_{3}^{2}+\ldots+(-1)^{n}(n+1) C_{n}^{2} \\ = & {\left[C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-C_{3}^{2}+\ldots+(-1)^{n} C_{n}^{2}\right] } \\ & -\left[C_{1}^{2}-2 C_{2}^{2}+3 C_{3}^{2}-\ldots+(-1)^{n} n C_{n}^{2}\right] \\ = & (-1)^{n / 2} \frac{n !}{(\frac{n}{2}) ! (\frac{n}{2}) !}-(-1)^{\frac{n}{2}-1} \frac{n}{2} \frac{n !}{(\frac{n}{2}) ! (\frac{n}{2}) !} \\ = & (-1)^{n / 2} \frac{n !}{(\frac{n}{2}) ! (\frac{n}{2}) !} (1+\frac{n}{2}) \end{aligned} $
$\therefore \frac{2 (\frac{n}{2}) ! (\frac{n}{2}) !}{n !}\left[C_{0}^{2}-2 C_{1}^{2}+3 C_{2}^{2}-\ldots+(-1)^{r}(n+1) C_{n}^{2}\right]$
$=\frac{2 (\frac{n}{2}) ! (\frac{n}{2}) !}{n !}(-1)^{n / 2} \frac{n !}{(\frac{n}{2}) ! (\frac{n}{2}) !} \frac{(n+2)}{2}=(-1)^{n / 2}(n+2)$
