Prove that (21/4 .41/8 . 81/16. 161/32….∞) = 2.

Let us consider the LHS

Now,

\( 2^{\dfrac{1}{4}} + \dfrac{2}{8} + \dfrac{3}{16} + \dfrac{1}{8} + …∞\)

So let us consider 2^x, where x = \( \dfrac{1}{4}+ \dfrac{2}{8} + \dfrac{3}{16} + \dfrac{1}{8} + …∞\)........ (1)

Multiply both sides of the equation with \( \dfrac{1}{2}\), we get

\(\dfrac{ x}{2} = \dfrac{1}{2} ( \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{16} + \dfrac{1}{8} + … ∞)\)

= \( \dfrac{1}{8} +\dfrac{2}{16} + \dfrac{3}{32} + … + ∞\) …. (2)

Now, subtract (2) from (1) we get,

\( x-\dfrac{ x}{2} = \dfrac{1}{2} ( \dfrac{1}{4} + \dfrac{2}{8} + \dfrac{3}{16} + \dfrac{1}{8} + … ∞)\) –

 \( \dfrac{1}{8} +\dfrac{2}{16} + \dfrac{3}{32} + … + ∞\)

By grouping similar terms,

\( x = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + … ∞\)

Where, a = \( \dfrac{1}{2}\), r =\(\dfrac{ (\dfrac{1}{4}) }{ (\dfrac{1}{2})} \)= \( \dfrac{1}{2}\)

By using the formula,

S_∞ = \( \dfrac{a}{(1 – r)}\)

= \(\dfrac{ (\dfrac{1}{2}) }{ (\dfrac{1}{2})}\)

= 1

From equation (1), 2^x = 2¹ = 2 = RHS

Hence proved.