A G.P has first term and common ratio both equal to \displaystyle \alpha ,\alpha >1.Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms , find the exact value of \displaystyle \alpha .
Hence evaluate \displaystyle {{\log }_{3}}(\frac{3}{2}{{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+...+{{\alpha }^{{58}}}) giving your answer in the form of \displaystyle A-{{\log }_{3}}B , where A and B are positive integers to be determined.
Solution
\displaystyle a=r=\alpha \ \ \ ,\ \ \ \alpha >1
\displaystyle {{S}_{{12}}}=28\ {{S}_{6}}
\displaystyle \frac{{a({{r}^{{12}}}-1)}}{{r-1}}=28\times \frac{{a({{r}^{6}}-1)}}{{r-1}}
\displaystyle \ \ \ \ \ \ {{r}^{6}}+1=28
\displaystyle \ \ \ \ \ \ \ \ \ \ {{r}^{6}}\ =27
\displaystyle \ \ \ \ \ \ \ \ \ \ r\ \ =\sqrt{3}\ \ \ \ (\because \alpha >1)
\displaystyle \ \ \ \ \ \ \ \therefore \alpha =\sqrt{3}
\displaystyle {{\log }_{3}}(\frac{3}{2}{{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+...+{{\alpha }^{{58}}})
\displaystyle ={{\log }_{3}}(\frac{9}{2}+{{3}^{2}}+{{3}^{3}}+{{3}^{4}}+...+{{3}^{{29}}})
\displaystyle ={{\log }_{3}}(\frac{9}{2}+\frac{{9({{3}^{{28}}}-1)}}{2})
\displaystyle ={{\log }_{3}}\frac{{{{3}^{{30}}}}}{2}
\displaystyle ={{\log }_{3}}{{3}^{{30}}}-{{\log }_{3}}2
\displaystyle =30-{{\log }_{3}}2
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