Binomial Theorem 2 Question 12
15.
Prove that $\frac{3 !}{2(n+3)}=\sum_{r=0}^{n}(-1)^{r} (\frac{{ }^{n} C_{r}}{{ }^{r+3} C_{r}})$
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Solution:
- $\sum_{r=0}^{n}(-1)^{r} \frac{{ }^{n} C_{r}}{{ }^{r+3} C_{r}}$
$ =\sum_{r=0}^{n}(-1)^{r} \frac{n ! \cdot 3 !}{(n-r) !(r+3) !}=3 ! \sum_{r=0}^{n}(-1)^{r} \frac{n !}{(n-r) !(r+3) !} $
$ =\frac{3 !}{(n+1)(n+2)(n+3)} \sum_{r=0}^{n} \frac{(-1)^{r} \cdot(n+3) !}{(n-r) !(r+3) !} $
$=\frac{3 !}{(n+1)(n+2)(n+3)} \cdot \sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n+3} C_{r+3}$
$=\frac{3 !(-1)^{3}}{(n+1)(n+2)(n+3)} \sum_{s=3}^{n+3}(-1)^{s} \cdot{ }^{n+3} C_{3}$
$ =\frac{-3 !}{(n+1)(n+2)(n+3)}( \sum_{s \doteq 0}^{n+3}(-1)^{s}{ }^{n+3} C_{s}){-{ }^{n+3} C_{0}+{ }^{n+3} C_{1}-{ }^{n+3} C_{2}} $
$=\frac{-3!}{(n+1)(n+2)(n+3)} \quad( 0-1+(n+3)-\frac{(n+3)(n+2)}{2 !}) $
$=\frac{-3 !}{(n+1)(n+2)(n+3)} \cdot \frac{(n+2)(2-n-3)}{2}=\frac{3 !}{2(n+3)}$
