Sequence And Series Question 304
Question: If $ n $ be odd or even, then the sum of $ n $ terms of the series $ 1-2+ $ $ 3- $ $ 4+5-6+…… $ will be
Options:
A) $ -\frac{n}{2} $
B) $ \frac{n-1}{2} $
C) $ \frac{n+1}{2} $
D) $ \frac{2n+1}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given series $ S=1-2+3-4+5-6……… $ Case I. If $ n $ is odd, say $ 2m+1 $ In this case, the number of positive terms $ =\frac{1}{2}(n+1)=\frac{1}{2}(2m+1+1)=(m+1) $ and the number of negative terms $ =(2m+1)-(m+1)=m $ Then sum $ =[1+3+5+………upto,(m+1)\ \text{terms }] $ $ -[2+4+6…….upto\ m\ terms] $ $ =\frac{1}{2}(m+1)[2+(m+1-1)2]-\frac{m}{2}[4+(m-1)2] $ $ =(m+1)(m+1-m)=m+1=\frac{1}{2}(n+1) $ . Case II. If $ n $ is even Sum $ =( 1+3+5……upto\ \frac{n}{2},terms ) $ $ -( 2+4+6….upto,\frac{n}{2}terms ) $ $ =\frac{1}{2}.\ \frac{n}{2}[ 2+( \frac{n}{2}-1 )2 ]-\frac{1}{2}.\frac{n}{2}[ 4+( \frac{n}{2}-1 )2 ] $ $ =\frac{1}{4}n[n-(n+2)]=-\frac{n}{2} $ . Trick: Put $ n=\ 3,,4 $ $ S_1=2,\ S_3=-,2, $ which the option (a) and (c) give for $ n=3,4 $ .
BETA
