If 0 < x < 1 and y = 1/2x² + 2/3x³ + 3/4x⁴ + ...., then the value of e 1 + y at x = 1/2 is :

Solution

Maths sir · 4 steps

Split each coefficient as 1 - 1/n: you get a GP minus a log series, and both have known sums.

1. Split the coefficient. The general term is (n-1)/nxⁿ = xⁿ - xⁿ/n for n ≥ 2. So y is a geometric series minus a logarithmic series.
2. Sum each part. The GP gives x² + x³ + … = x²/(1-x). Since -ln(1-x) = x + x²/2 + x³/3 + …, the log part is x²/2 + x³/3 + … = -ln(1-x) - x.

   y = x²/(1-x) + x + ln(1-x) = x/(1-x) + ln(1-x)

3. Put in x. At x = 1/2, x/(1-x) = 1 and ln 1/2 = -ln 2. So y = 1 - ln 2 and 1 + y = 2 - ln 2.
4. So. e¹⁺ʸ = e²⁻ˡⁿ² = e²/eˡⁿ² = 1/2e², which is option (a).

The idea to keep. When the coefficients look like (n-1)/n or n/(n+1), split them into 1 ± 1/n. One piece is a GP and the other is the ln(1-x) series.

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