The value of sumlimits_{r = 1}^n {log left( { | vedclass

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Question

(c) The given series is

$\log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \log \left( {\frac{{{a^

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) The given series is

$\log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \log \left( {\frac{{{a^4}}}{{{b^3}}}} \right) + ...... + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$

This is an $A.P.$ with first term $\log a$

and the common difference $\log \left( {\frac{{{a^2}}}{b}} \right) - \log a = \log \left( {\frac{a}{b}} \right)$

Therefore the sum of $n$ terms is

$\frac{n}{2}\left[ {\log a + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)} \right] = \frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$.

Trick : Check for $n = 1,\;2$.

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

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