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The sum of n terms of the series \(\displaystyle\sum_{x=1}^n\) log \(\frac{2^x}{3^{x-1}}\) is :

(a) log \(\bigg(\frac{3^{n-1}}{2^{n+1}}\bigg)^{\frac{n}{2}}\)

(b) log \(\bigg(\frac{2^{n-1}}{3^{n+1}}\bigg)^{\frac{n}{2}}\) 

(c) log \(\bigg(\frac{3^{n+1}}{2^{n-1}}\bigg)^{\frac{n}{2}}\) 

(d) log \(\bigg(\frac{2^{n+1}}{3^{n-1}}\bigg)^{\frac{n}{2}}\) 

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(d) log \(\bigg(\frac{2^{n+1}}{3^{n-1}}\bigg)^{\frac{n}{2}}\) 

  \(\displaystyle\sum_{x=1}^n\) log \(\frac{2^x}{3^{x-1}}\) = log \(\big(\frac{2^1}{3^0}\big)\) + log \(\big(\frac{2^2}{3^1}\big)\) + log \(\big(\frac{2^3}{3^2}\big)\) + ..... + log \(\big(\frac{2^n}{3^{n-1}}\big)\)

= log\(\big(\) \(\frac{2^1}{3^0}\)\(\frac{2^2}{3^1}\).\(\frac{2^3}{3^2}\) . ........ \(\frac{2^n}{3^{n-1}}\big)\)

= log \(\big(\)\(\frac{2^{1+2+3+.....+n}}{3^{1+2+3+.....+(n+1)}}\big)\) = log \(\bigg[\frac{2^{\frac{n(n+1)}{2}}}{3^{\frac{n(n-1)}{2}}}\bigg]\) =  log \(\bigg[\frac{2^{n+1}}{3^{n-1}}\bigg]^{\frac{n}{2}}\) 

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