The value of $\mathop \sum \limits_{r = 1}^n \log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)$ is

$\frac{n}{2}\log \left( {\frac{{{a^n}}}{{{b^n}}}} \right)$

$\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^n}}}} \right)$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

The value of

\sum \limits_{r = 1}^n \log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)

View answer and explanation

Correct option: C

\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)

Option C (

\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)

) is correct because by applying logarithm properties and simplifying the summation, the expression is equivalent to the given option.

Practice All Chapter MCQs →

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\frac{3}{4}