Sequence And Series Question 121
Question: The sum 1(1!) + 2(2!) + 3(3!) + ….+n (n!) equals
[AMU 1999; DCE 2005]
Options:
A) $ 3,(n,!),+,n-3 $
B) $ (n+1)!,-,(n-1)! $
C) $ (n+1),!,-1 $
D) $ 2,(n,!)-2n-1 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ S_{n}=1(1!)+2(2!)+3(3!)+…..+n(n!) $ = $ (2-1)(1!)+(3-1)(2!)+(4-1)(3!)+….. $ $ +(n+1)-1 $ = $ (2.1!-1!)+(3.2!-2!)+(4.3!-3!)+…. $ $ +[(n+1)(n!)-(n!)] $ = $ (2!-1!)+(3!-2!)+(4!-3!)+….+[(n+1)!-(n)!] $ = $ (n+1)!-1! $ .
BETA
