Find the sum of n terms of the series loga+loga2b+loga3b2+loga4b3+__________ to n terms.
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Solution
Tn−Tn−1=loganbn−1−logan−1bn−2 =loganbn−1⋅bn−2an−1=logab=constant as it is independent of n. Hence the given series is an A.P. whose A=loga and D=logab. ∴S=n2[2loga+(n−1)logab] =n2[nlogab+2loga−(loga−logb)] =n2[nlogab+logab].