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Question

Sum to n terms of the series log m+log m2n+log m3n2+log m4n3 is

A
log(mn+1nn1)n2
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B
log(nn1mn+1)n2
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C
log(mnnn)n2
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D
log(m1nn1m)n2
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Solution

The correct option is A log(mn+1nn1)n2
log (m × m2 × m3 × .........mn)log(1 × n × n2 × ......n(n1)) = log[m(n+1)n(n1)](n/2)

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