Gate math book

Question

Find the sum of n terms of the series

$log a+ log \frac{a^{2}}{b} + log \frac{a^{3}}{b^{2}}+ ...$ to n terms

Comments

I am getting $\frac{n}{2}\left [ n log ab+ log \frac{a}{b}\right ]$ but answer given is different

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is this is the answer : $n*log(a^{n/2}/b^{n+1})$

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No it is incorrect

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both answer are correct how can i give 2 best answer ?

Answer 1

.....

Comments

You are correct

Answer 2

$\begin{align*} &\;\log a + \log \left ( \frac{a^2}{b} \right ) + \log \left ( \frac{a^3}{b^2} \right ) + \log \left ( \frac{a^4}{b^3} \right ) + \dots + \log \left ( \frac{a^n}{b^{n-1}} \right ) \\ &=\sum_{k=1}^{n} \log \left ( \frac{a^k}{b^{k-1}} \right ) \\ &=\sum_{k=1}^{n}\log a^k - \sum_{k=1}^{n} \log b^{k-1} \\ &=\log a \cdot \sum_{k=1}^{n}k - \log b \cdot \sum_{k=1}^{n}(k-1) \\ &=\log a \cdot \left [ \frac{n(n+1)}{2} \right ] - \log b \cdot \left [ \frac{n(n-1)}{2} \right ] \\ &= \log \left [ \frac{a^{\frac{n(n+1)}{2}}}{b^{\frac{n(n-1)}{2}}} \right ] \\ \end{align*}$

Comments

I was writing : it took late : already answered :)

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:D i have another method to solve this how many best answers possible :D

Answer 3

hope it might help........

Comments

you are also correct