수학1_여러 가지 수열_시그마합공식_난이도 중

다음은 등식 $\sum \limits_{k=1}^{n} k(k+1)(k+2)(k+3)=\dfrac{n(n+1)(n+2)(n+3)(n+4)}{5}$ 가 성립함을 증명한 것이다.

$\sum \limits_{k=1}^{n} k(k+1)(k+2)(k+3)$ $=\sum \limits_{k=1}^{n} (가)$

$=4! \left \{ \dfrac{4!}{4! \times 0!} + \dfrac{5!}{4! \times 1!} + \cdots + \dfrac{(n+3)!}{4! \times (n-1)!} \right \}$ $=4! \cdot \sum \limits_{k=1}^{n} (나)$

$= 4! \cdot  (다)$ $=\dfrac{n(n+1)(n+2)(n+3)(n+4)}{5}$ 

위 과정에서 (가), (나), (다)에 알맞은 것은?

	(가)	(나)	(다)
①	$\dfrac{(k+2)!}{(k-1)!}$	$_{k+2}{\rm C}_3$	$_{n+3}{\rm C}_4$
②	$\dfrac{(k+2)!}{(k-1)!}$	$_{k+2}{\rm C}_3$	$_{n+4}{\rm C}_5$
③	$\dfrac{(k+3)!}{(k-1)!}$	$_{k+2}{\rm C}_3$	$_{n+3}{\rm C}_4$
④	$\dfrac{(k+3)!}{(k-1)!}$	$_{k+3}{\rm C}_4$	$_{n+3}{\rm C}_4$
⑤	$\dfrac{(k+3)!}{(k-1)!}$	$_{k+3}{\rm C}_4$	$_{n+4}{\rm C}_5$

 

정답 ⑤