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

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Question

(C) The given series is $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} = \log a + \log \left( {\frac{{{a^2}}}{b}} \right

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given series is $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} = \log a + \log \left( {\frac{{{a^2}}}{b}} \right) + \log \left( {\frac{{{a^3}}}{{{b^2}}}} \right) + \dots + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$.This is an arithmetic progression $(A.P.)$ where the first term $A = \log a$ and the last term $L = \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right)$.The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(A + L)$.Substituting the values,we get $S_n = \frac{n}{2} \left[ \log a + \log \left( {\frac{{{a^n}}}{{{b^{n - 1}}}}} \right) \right]$.Using the property $\log x + \log y = \log(xy)$,we have $S_n = \frac{n}{2} \log \left( a \cdot \frac{{{a^n}}}{{{b^{n - 1}}}} \right) = \frac{n}{2} \log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)$.

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

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