Question

Prove that : 

(2 ^1/4 .4^1/8 . 8^1/16. 16^1/32….∞) = 2.

Answer 1

Let,  S = 2^{\(\frac{1}{4}\)}, 2^{\(\frac{2}{3}\)}, 2^{\(\frac{3}{16}\)}, 2^{\(\frac{1}{8}\)},

Using the properties of exponents: 

The above term can be written as:

Denoting the terms in power with x, 

We have

Clearly, we observe that x is neither possessing any common ratio or any common difference. But if you observe carefully you can see that numerator is possessing an AP and denominator of various terms are in GP 

Many of similar problems are solved using the method of difference approach as solved below:

As x = \(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + ..... \infty\)...... Equation 1

Multiply both sides of the equation with 1/2,we have-

Clearly, we have a progression with common ratio = 1/2 

∴ it is a Geometric progression

Sum of infinite GP = \(\frac{a}{1-r}\) ,where a is the first term and r is the common ratio. 

Note: We can only use the above formula if |r|<1

Clearly, a = \(\frac{1}{2}\) and r = \(\frac{1}{2}\)

⇒ x = \(\cfrac{\frac{1}{2}}{1-(\frac{1}{2})}\) = 1

From equation 1 we have, 

S = 2^x = 2¹ = 2 = RHS