SATHEE: Binomial Theorem 2 Question 12

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:

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

$= \sum_{r = 0}^{n} (- 1)^{r} \frac{n! \cdot 3!}{(n - r)! (r + 3)!} = 3! \sum_{r = 0}^{n} (- 1)^{r} \frac{n!}{(n - r)! (r + 3)!}$

$= \frac{3!}{(n + 1) (n + 2) (n + 3)} \sum_{r = 0}^{n} \frac{(- 1)^{r} \cdot (n + 3)!}{(n - r)! (r + 3)!}$

$= \frac{3!}{(n + 1) (n + 2) (n + 3)} \cdot \sum_{r = 0}^{n} (- 1)^{r} \cdot^{n + 3} C_{r + 3}$

$= \frac{3! (- 1)^{3}}{(n + 1) (n + 2) (n + 3)} \sum_{s = 3}^{n + 3} (- 1)^{s} \cdot^{n + 3} C_{3}$

$= \frac{- 3!}{(n + 1) (n + 2) (n + 3)} (\sum_{s ≐ 0}^{n + 3} (- 1)^{s}^{n + 3} C_{s}) -^{n + 3} C_{0} +^{n + 3} C_{1} -^{n + 3} C_{2}$

$= \frac{- 3!}{(n + 1) (n + 2) (n + 3)} (0 - 1 + (n + 3) - \frac{(n + 3) (n + 2)}{2!})$

$= \frac{- 3!}{(n + 1) (n + 2) (n + 3)} \cdot \frac{(n + 2) (2 - n - 3)}{2} = \frac{3!}{2 (n + 3)}$

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