If the value of (1 + 2/3 + 6/3^2 + 10/3^3 + ...... upto ∞)^(log_(0.25) (1/3 + 1/3^2 + 1/3^3 + ..... upto ∞)) is l then l^2

Question

If the value of

\(\left( 1 + \frac{2}{3} +\frac{6}{3^2} + \frac{10}{3^3} + ...... upto \,\infty\right)\)^{\(log_{(0.25)}\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ..... upto \, \infty\right)\)} 

is l, then l² is equal to ............. .

Answer 1

Correct  answer is 3

\(\frac{2S}{2} = \frac{4}{3} + \frac{4}{3^2}+ \frac{4}{3^3} + .....\)

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

Now ℓ = (3)^{log _0.25 \(\left(\frac{1/3}{1-1/3}\right)\)} 

ℓ = 3^{log((1/4))\(\left(1/2\right)\)} = 3^1/2 =\(\sqrt{3}\)

\(\Rightarrow\) ℓ² = 3