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Question

Sum of n terms of the series log a+log(a2b)+log(a3b2)+log(a4b3)+ is:

A
log{anbn1}n/2
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B
log{an+1bn1}n/2
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C
log{anbn}n/2
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D
log{an+1bn}n/2
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Solution

The correct option is B log{an+1bn1}n/2
log a+loga2b+loga3b2+loga4b3++loganbn1
=log(a.a2b.a3b2.a4b3annn1)
=log[a(1+2+3+4++n)b(1+2+3+4++(n1))]
=log[an(n+1)2b(n1)n2]=log[a(n+1)b(n1)]n/2

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