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Question
Let rth term of a series is given by, Tr=r1−3r2+r4.
Then limn→∞∑nr=1Tr is
Let rth term of a series is given by, Tr=r1−3r2+r4.
Then limn→∞∑nr=1Tr is
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Solution
The correct option is C −12
Tr=r(r2−1)2−r2=r(r2−1+r)(r2−1−r)
Tr=12(r2+r−1)−(r2−r−1)(r2−1+r)(r2−1−r)
Tr=12[1(r2−1−r)−1(r2−1+r)]
T1=12[−1−11]
T2=12[11−15]
T3=12[15−111]
Tn=12[1(n2−1−n)−1(n2−1+n)]
Adding all we getSn=12[−1−1(n2−1−n)]
limn→∞∑nr=1Tn=limn→∞12[−1−1(n2−1−n)]=−12(1−0)=−12
Tr=r(r2−1)2−r2=r(r2−1+r)(r2−1−r)
Tr=12(r2+r−1)−(r2−r−1)(r2−1+r)(r2−1−r)
Tr=12[1(r2−1−r)−1(r2−1+r)]
T1=12[−1−11]
T2=12[11−15]
T3=12[15−111]
Tn=12[1(n2−1−n)−1(n2−1+n)]
Adding all we get
Sn=12[−1−1(n2−1−n)]
limn→∞∑nr=1Tn=limn→∞12[−1−1(n2−1−n)]=−12(1−0)=−12
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