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) $(-1)^{n}(n+1)$
(1986, 2M) (1997C, 5M)
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Answer:
Correct Answer: 10. (a)
Solution:
- We have,
$$ \begin{aligned} C _0^{2}-2 & C _1^{2}+3 C _2^{2}-4 C _3^{2}+\ldots+(-1)^{n}(n+1) C _n^{2} \\ = & {\left[C _0^{2}-C _1^{2}+C _2^{2}-C _3^{2}+\ldots+(-1)^{n} C _n^{2}\right] } \\ & -\left[C _1^{2}-2 C _2^{2}+3 C _3^{2}-\ldots+(-1)^{n} n C _n^{2}\right] \\ = & (-1)^{n / 2} \frac{n !}{\frac{n}{2} ! \frac{n}{2} !}-(-1)^{\frac{n}{2}-1} \frac{n}{2} \frac{n !}{\frac{n}{2} ! \frac{n}{2} !} \\ = & (-1)^{n / 2} \frac{n !}{\frac{n}{2} ! \frac{n}{2} !} 1+\frac{n}{2} \end{aligned} $$
$\therefore \frac{2 \frac{n}{2} ! \frac{n}{2} !}{n !}\left[C _0^{2}-2 C _1^{2}+3 C _2^{2}-\ldots+(-1)^{r}(n+1) C _n^{2}\right]$
$=\frac{2 \frac{n}{2} ! \frac{n}{2} !}{n !}(-1)^{n / 2} \frac{n !}{\frac{n}{2} ! \frac{n}{2} !} \frac{(n+2)}{2}=(-1)^{n / 2}(n+2)$
