A) \[-\frac{n}{2}\]
B) \[\frac{n-1}{2}\]
C) \[\frac{n+1}{2}\]
D) \[\frac{2n+1}{2}\]
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 \[=[1+3+5+.........\text{upto}\,(m+1)\ \text{terms }\!\!]\!\!\text{ }\]\[-[2+4+6.......\text{upto}\ m\ \text{terms}]\] \[=\frac{1}{2}(m+1)[2+(m+1-1)2]-\frac{m}{2}[4+(m-1)2]\] \[=(m+1)(m+1-m)=m+1=\frac{1}{2}(n+1)\]. Case II. If \[n\] is even Sum \[=\left( 1+3+5......\text{upto}\ \frac{n}{2}\,\text{terms} \right)\]\[-\left( 2+4+6....\text{upto}\,\frac{n}{\text{2}}\text{terms} \right)\] \[=\frac{1}{2}.\ \frac{n}{2}\left[ 2+\left( \frac{n}{2}-1 \right)2 \right]-\frac{1}{2}.\frac{n}{2}\left[ 4+\left( \frac{n}{2}-1 \right)2 \right]\] \[=\frac{1}{4}n[n-(n+2)]=-\frac{n}{2}\]. Trick: Put \[n=\ 3,\,4\] \[{{S}_{1}}=2,\ {{S}_{3}}=-\,2,\] which the option (a) and (c) give for \[n=3,4\].
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