If 0 < x < 1, then 3/2x² + 5/3x³ + 7/4x⁴ + ....., is equal to :

Solution

Maths sir · 4 steps

Split each coefficient as 2 minus a fraction: one part is a GP and the other is the log series.

1. General term. The nth term is (2n+1)/(n+1)xⁿ⁺¹ for n = 1, 2, 3, … Since 2n+1 = 2(n+1) - 1, the coefficient is 2 - 1/(n+1).

   S = 2Σₙ₌₁^∞ xⁿ⁺¹ - Σₙ₌₁^∞xⁿ⁺¹/(n+1)

2. Two known sums. The first sum is a GP: x² + x³ + … = x²/(1-x). The second is x²/2 + x³/3 + …, which is the series for -logₑ(1-x) without its first term x. So it equals -logₑ(1-x) - x.
3. Combine. S = 2x²/(1-x) + x + logₑ(1-x). Put the first two terms over one denominator: (2x² + x - x²)/(1-x) = x(1+x)/(1-x).
4. So. S = x((1+x)/(1-x)) + logₑ(1-x), which is option (a).

The idea to keep. When the coefficient is a fraction like (an+b)/(n+c), divide it out into a constant plus k/(n+c). The constant part gives a GP, and the 1/(n+c) part gives a log series.

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