If 0 < a, b < 1, and tan⁻¹a + tan⁻¹b = π/4, then the value of
(a + b) - ( (a² + b²)/2 ) + ( (a³ + b³)/3 ) - ( (a⁴ + b⁴)/4 ) + ..... is :

Solution

Maths sir · 4 steps

Split the series into two log series, then use the tan inverse condition to get (1+a)(1+b).

1. Split the series. Regroup the terms as (a - a²/2 + a³/3 - …) + (b - b²/2 + b³/3 - …). Both brackets converge because 0 < a, b < 1, so each one is the series for logₑ(1+x).

   S = logₑ(1+a) + logₑ(1+b) = logₑ[(1+a)(1+b)]

2. Use the condition. Since ab < 1, tan⁻¹ a + tan⁻¹ b = tan⁻¹(a+b)/(1-ab) = π/4. So (a+b)/(1-ab) = 1, which gives a + b + ab = 1.
3. Product of brackets. (1+a)(1+b) = 1 + a + b + ab = 1 + 1 = 2.
4. So. S = logₑ 2, which is option (a).

The idea to keep. Alternating powers divided by n means a logₑ(1+x) series. Split it by variable, add the logs, and turn the given condition into the product inside the log.

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