Binomial Theorem 2 Question 12
15. Prove that $\frac{3 !}{2(n+3)}=\sum _{r=0}^{n}(-1)^{r} \frac{{ }^{n} C _r}{{ }^{r+3} C _r}$
$(2002,1 M)$ maximum when $m$ is equal to
(d) 20
(a) 5
(b) 10
(c) 15
is equal to
(2000, 2M)
(a) $\begin{aligned} & n+1 \ & r-1\end{aligned}$
(b) $2 \begin{aligned} & n+1 \ & r+1\end{aligned}$
(c) $2 \stackrel{n+2}{r}$
(d) $\begin{gathered}n+2 \ r\end{gathered}$
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Solution:
- $\sum _{r=0}^{n}(-1)^{r} \frac{{ }^{n} C _r}{{ }^{r+3} C _r}$
$$ =\sum _{r=0}^{n}(-1)^{r} \frac{n ! \cdot 3 !}{(n-r) !(r+3) !}=3 ! \sum _{r=0}^{n}(-1)^{r} \frac{n !}{(n-r) !(r+3) !} $$
$$ =\frac{3 !}{(n+1)(n+2)(n+3)} \sum _{r=0}^{n} \frac{(-1)^{r} \cdot(n+3) !}{(n-r) !(r+3) !} $$
$=\frac{3 !}{(n+1)(n+2)(n+3)} \cdot \sum _{r=0}^{n}(-1)^{r} \cdot{ }^{n+3} C _{r+3}$
$=\frac{3 !(-1)^{3}}{(n+1)(n+2)(n+3)} \sum _{s=3}^{n+3}(-1)^{s} \cdot{ }^{n+3} C _3$
$$ =\frac{-3 !}{(n+1)(n+2)(n+3)} \sum _{s \doteq 0}^{n+3}(-1)^{s}{ }^{n+3} C _s $$
$$ \begin{aligned} & =\frac{-{ }^{n+3} C _0+{ }^{n+3} C _1-{ }^{n+3} C _2}{(n+1)(n+2)(n+3)} \quad 0-1+(n+3)-\frac{(n+3)(n+2)}{2 !} \\ & =\frac{-3 !}{(n+1)(n+2)(n+3)} \cdot \frac{(n+2)(2-n-3)}{2}=\frac{3 !}{2(n+3)} \end{aligned} $$
