Sequence And Series Question 156
Question: The sum to $ n $ terms of $ (2n-1)+2,(2n-3) $ $ +3,(2n-5)+….. $ is
[AMU 2001]
Options:
A) $ (n+1),(n+2),(n+3)/6 $
B) $ n,(n+1),(n+2)/6 $
C) $ n,(n+1),(2n+3), $
D) $ n,(n+1),(2n+1)/6 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=(2n-1)+2(2n-3)+3(2n-5)+…. $ $ S=[2n+2.2n+3.2n+……+n.2n]- $ $ [1+2.3+3.5+….+n.(2n-1)] $ Let, $ S_1=2n(1+2+3+….+n) $ = $ \frac{2n.n(n+1)}{2}=n^{2}(n+1) $ and $ S_2=1+2.3+3.5+…..+n.(2n-1) $ $ T_{n}=n(2n-1)=2n^{2}-n $
$ \therefore S_2=\sum (2n^{2}-n) $ $ =2\sum (n^{2})-\sum (n) $ $ =\frac{2n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2} $
$ \therefore S=S_1-S_2=n^{2}(n+1)-\frac{2n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} $ $ =n,(n+1)[ n-\frac{2n+1}{3}+\frac{1}{2} ] $ $ =n,(n+1)[ \frac{6n-4n-2+3}{6} ] $ $ =\frac{n,(n+1)(2n+1)}{6} $ .
BETA
