If the value of
( 1 + 2/3 + 6/3² + 10/3³ + ....upto ∞ )^(log _((0.25))( 1/3 + 1/3² + 1/3³ + ....upto ∞ ))
is l, then l² is equal to _______________.

Solution

Maths sir · 4 steps

The base is an AGP and the exponent is a GP. Sum each one, and the power turns into a square root.

1. Spot the AGP. Leave the first term 1 aside. The numerators 2, 6, 10, … form an AP with common difference 4, and the denominators 3, 3², 3³, … form a GP with ratio 1/3. So the rest of the base is an arithmetico-geometric series. Call it S.

   S - S/3 = 2/3 + 4/3² + 4/3³ + … = 2/3 + (4/9)/(1 - 1/3) = 2/3 + 2/3 = 4/3

2. Sum the base. From the line above, 2/3S = 4/3, so S = 2. Adding back the leading 1 makes the base 1 + 2 = 3.
3. Sum the exponent. The series 1/3 + 1/3² + … is a GP with sum (1/3)/(1 - 1/3) = 1/2. Since 0.25 = (1/2)², we get log_(0.25) 1/2 = 1/2.
4. So. l = 3^(1/2) = √3, and so l² = 3.

The common slip: 2. Forgetting the leading 1 makes the base 2 instead of 3, so l = √2 and l² = 2.

The idea to keep. For an AGP with ratio r, find S - rS. What is left is a plain GP. Also write the base of a log as a power of the argument before evaluating it, like 0.25 = (1/2)².

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