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First Fundamental Theorem of Calculus

The value of ...

Question

The value of $\sum_{n = 1}^{m} log \frac{a^{2 n - 1}}{b^{m - 1}} (a \neq 0; b \neq 0, 1)$ is equal to

A

m log \frac{a^{2 m}}{b^{m - 1}}

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B

log \frac{a^{2 m}}{b^{m - 1}}

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C

\frac{m}{2} log \frac{a^{2 m}}{b^{2 m - 2}}

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D

\frac{m}{2} log \frac{a^{2 m}}{b^{m + 1}}

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Solution

The correct option is C $\frac{m}{2} log \frac{a^{2 m}}{b^{2 m - 2}}$
Let $S = \sum_{n = 1}^{m} log \frac{a^{2 n - 1}}{b^{m - 1}}$
$\Rightarrow S = \sum_{n = 1}^{m} (log a^{2 n - 1} - log b^{m - 1}) = log a \sum_{n = 1}^{m} (2 n - 1) - (m - 1) log b \sum_{n = 1}^{m}$
$\Rightarrow S = - m (m - 1) log b + log a {2 \sum_{n = 1}^{m} n - \sum_{n = 1}^{m}}$
$\Rightarrow S = - m (m - 1) log b + log a [m (m + 1) - m]$
$\Rightarrow S = \frac{m}{2} [2 m log a - (2 m - 2) log b]$
$\Rightarrow S = \frac{m}{2} log \frac{a^{2 m}}{b^{2 m - 2}}$
Ans: C

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Similar questions

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B M

We are giving the concept of A.M of

m^{t h}

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then the correct statement is:

Question 20

For non-zero numbers $a$ and $b$ , ${(\frac{a}{b})}^{m} \div {(\frac{a}{b})}^{n},$ where $m > n,$ is equal to:
(a) ${(\frac{a}{b})}^{m n}$
(b) ${(\frac{a}{b})}^{m + n}$
(c) ${(\frac{a}{b})}^{m - n}$
(d) ${({(\frac{a}{b})}^{m})}^{n}$

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