If the value of left(1+frac{2}{3}+frac{6}{3^{ | vedclass

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(B) Let $S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \ldots \infty$.Dividing by $3$,we get $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}}

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$3$

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(B) Let $S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \ldots \infty$.Dividing by $3$,we get $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{6}{3^{3}} + \ldots \infty$.Subtracting the two equations: $S - \frac{S}{3} = 1 + \frac{1}{3} + \frac{4}{3^{2}} + \frac{4}{3^{3}} + \ldots \infty$.$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{4}{3^{2}}(1 + \frac{1}{3} + \ldots \infty) = \frac{4}{3} + \frac{4}{3^{2}} \left( \frac{1}{1 - 1/3} ight) = \frac{4}{3} + \frac{4}{9} \left( \frac{3}{2} ight) = \frac{4}{3} + \frac{2}{3} = 2$.Thus,$S = 2 	imes \frac{3}{2} = 3$.Now,the exponent is $\log_{0.25} \left( \frac{1/3}{1 - 1/3} ight) = \log_{1/4} \left( \frac{1/3}{2/3} ight) = \log_{1/4} \left( \frac{1}{2} ight) = \log_{(1/2)^{2}} (1/2) = \frac{1}{2}$.Therefore,$l = 3^{1/2} = \sqrt{3}$.Hence,$l^{2} = 3$.

If the value of $\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots \text{ to } \infty\right)^{\log_{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \text{ to } \infty\right)}$ is $l$,then $l^{2}$ is equal to $......$

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