Let the rth term, tr of a series is given by tr=r1+r2+r4. Then limn→∞∑nr=1tr is
A
14
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B
1
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C
12
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D
13
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Solution
The correct option is B12 Given, tr=r1+r2+r4 =12.2r(r2+1)2−r2=12(1r(r−1)−1r(r+1)+1) Let f(r)=1r(r+1)+1 Therefore, ∑nr=1tr=∑nr=112{f(r)−f(r−1)} =12{f(1)−f(n−1)}=12{1−1(n+1)(n+1)} Hence, limn→∞∑nr−1tr=limn→∞12{1−1(n+1)(n+1)}=12