The product 2^(1/4).4^(1/16).8^(1/48).16^(1/128) ... to ∞ is equal to :

Solution

Maths sir · 3 steps

Write every factor as a power of 2. Then the infinite product becomes an infinite sum of exponents.

1. Same base. The nth factor is (2ⁿ)^(1/(n · 2ⁿ⁺¹)), since the denominators are 4 = 1 × 4, 16 = 2 × 8, 48 = 3 × 16 and 128 = 4 × 32. So 4^(1/16) = 2^(2/16) = 2^(1/8), 8^(1/48) = 2^(3/48) = 2^(1/16) and 16^(1/128) = 2^(4/128) = 2^(1/32).
2. Add the exponents. When powers of the same base are multiplied, their exponents add. The exponents 1/4, 1/8, 1/16, 1/32, … form an infinite GP with first term 1/4 and ratio 1/2.

   1/4 + 1/8 + 1/16 + … = (1/4)/(1 - 1/2) = 1/2

3. So. The product equals 2^(1/2), which is option (b).

The idea to keep. An infinite product of powers is really an infinite sum in disguise. Bring every factor to one base, add the exponents, and the sum is usually a GP.

Answer from the NTA final answer key. Solution by PrepWiser.