Sum of n terms of the series loga+log(a2b)+log(a3b2)+log(a4b3)+⋯∞ is:
A
log{anbn−1}n/2
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B
log{an+1bn−1}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 Blog{an+1bn−1}n/2 loga+loga2b+loga3b2+loga4b3+⋯+loganbn−1 =log(a.a2b.a3b2.a4b3⋯annn−1) =log[a(1+2+3+4+⋯+n)b(1+2+3+4+⋯+(n−1))] =log[an(n+1)2b(n−1)n2]=log[a(n+1)b(n−1)]n/2