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Question

Let the rth term, tr of a series is given by tr=r1+r2+r4. Then limnnr=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 B 12
Given, tr=r1+r2+r4
=12.2r(r2+1)2r2=12(1r(r1)1r(r+1)+1)
Let f(r)=1r(r+1)+1
Therefore, nr=1tr=nr=112{f(r)f(r1)}
=12{f(1)f(n1)}=12{11(n+1)(n+1)}
Hence, limnnr1tr=limn12{11(n+1)(n+1)}=12

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