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Prove that :

(21/4 .41/8 . 81/16. 161/32….∞) = 2.

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Best answer

Let us consider the LHS

(21/4 .41/8 . 81/16. 161/32….∞)

This can be written as

21/4 . 22/8 . 23/16 . 21/8 … ∞

Now,

21/4 + 2/8 + 3/16 + 1/8 + …∞

So let us consider 2x, where x = 1/4 + 2/8 + 3/16 + 1/8 + … ∞ …. (1)

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

x/2 = 1/2 (1/4 + 2/8 + 3/16 + 1/8 + … ∞)

= 1/8 + 2/16 + 3/32 + … + ∞ …. (2)

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

x – x/2 = (1/4 + 2/8 + 3/16 + 1/8 + … ∞) – (1/8 + 2/16 + 3/32 + … + ∞)

By grouping similar terms,

x/2 = 1/4 + (2/8 – 1/8) + (3/16 – 2/16) + … ∞

x/2 = 1/4 + 1/8 + 1/16 + … ∞

x = 1/2 + 1/4 + 1/8 + 1/16 + … ∞

Where, a = 1/2, r = (1/4) / (1/2) = 1/2

By using the formula,

S = a/(1 – r)

= (1/2) / (1 – 1/2)

= (1/2) / ((2 - 1)/2)

= (1/2) / (1/2)

= 1

From equation (1), 2x = 21 = 2 = RHS

Hence proved.

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