Prove that $$(2^{\dfrac{1}{4}} .4^{\dfrac{1}{8}} . 8^{\dfrac{1}{16}}. 16^{\dfrac{1}{32}}….∞) = 2.$$

Asked by Sakshi | 1 year ago |  76

##### Solution :-

Let us consider the LHS

Now,

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

So let us consider 2x, 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), 2x = 21 = 2 = RHS

Hence proved.

Answered by Sakshi | 1 year ago

### Related Questions

#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

#### Find the two numbers whose A.M. is 25 and GM is 20.

Find the two numbers whose A.M. is 25 and GM is 20.

#### If a is the G.M. of 2 and 1/4 find a.

If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.

#### Find the geometric means of the following pairs of numbers

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a3b and ab3

(iii) –8 and –2

Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$.