The sum of the series $$S=$$∑r−$$1n$$ log⁡$$ar+1br$$−$$1$$ is

Correct option is 2n2log⁡an+3bn−1

Questions

The sum of the series $S = \sum_{r - 1}^{n} \log (\frac{a^{r + 1}}{b^{r - 1}})$ is

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n2log⁡an−1bn

n2log⁡an+3bn−1

n2log⁡an+2bn−1

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detailed solution

Correct option is B

S=∑r=1n [(r+1)log⁡a−(r−1)log⁡b].=n2[(2+n+1)log⁡a−(0+n−1)log⁡b]=n2log⁡an+3bn−1

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The sum of the series S=∑r−1n log⁡ar+1br−1 is

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The sum of the series $S = \sum_{r - 1}^{n} \log (\frac{a^{r + 1}}{b^{r - 1}})$ is