Let the rth term, tr, of a series is given by tr=r1+r2+r4. Then limn→∞n∑r=1tr is
A
14
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B
1
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C
12
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D
none of these
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Solution
The correct option is C12 As Tr=r1+r2+r4=122r1+r2+r4=12(1r2−r+1−1r2+r+1)=12(1r(r−1)+1−1r(r+1)+1) Then ∑nr=1Tr=∑nr=112(1r(r−1)+1−1r(r+1)+1)=12(1−1n(n+1)+1) Therefore, Ltn→∞∑nr=1Tr=Ltn→∞12(1−1n(n+1)+1)=12 Hence, option 'C' is correct.