# If the value of

Question:

If the value of

$\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots \ldots \text { upto } \infty\right)^{\log _{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \ldots \text { upto } \infty\right)}$

is $l$, then $l^{2}$ is equal to_________.

Solution:

$\ell=(\underbrace{1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}}_{\mathrm{s}}+\ldots)^{\log _{025}\left(\frac{1}{3}+\frac{1}{3^{2}}+\ldots\right)}$

$\mathrm{S}=1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots$

$\frac{\mathrm{S}}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{6}{3^{3}}+\ldots$

$\frac{2 S}{3}=1+\frac{1}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\ldots .$

$\frac{2 S}{3}=\frac{4}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\ldots$

$\mathrm{S}=\frac{3}{2}\left(\frac{4 / 3}{1-1 / 3}\right)=3$

Now $\ell=(3)^{\log _{025}\left(\frac{1 / 3}{1-1 / 3}\right)}$

$\ell=3^{\log _{(1 / 4)}\left(\frac{1}{2}\right)}=3^{1 / 2}=\sqrt{3}$

$\Rightarrow \ell^{2}=3$

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