The sum of the series 1/(1-3 · 1²+1⁴)+2/(1-3 · 2²+2⁴)+3/(1-3 · 3²+3⁴)+… up to 10 -terms is

Solution

Maths sir · 4 steps

Factor r⁴ - 3r² + 1 as a difference of squares, and the numerator r splits each term into two fractions.

1. Factor the denominator. Write r⁴ - 3r² + 1 = (r² - 1)² - r². As a difference of squares, this is (r² - r - 1)(r² + r - 1).
2. Split the term. The two factors differ by (r² + r - 1) - (r² - r - 1) = 2r. That is twice the numerator, so each term breaks into a difference.

   Tᵣ = r/(r² - r - 1)(r² + r - 1) = 1/2(1/(r² - r - 1) - 1/(r² + r - 1))

3. Telescope. Let f(r) = r² + r - 1. Then r² - r - 1 = f(r-1), so Tᵣ = 1/2(1/f(r-1) - 1/f(r)). Adding from r = 1 to 10, every middle fraction cancels and only 1/2(1/f(0) - 1/f(10)) is left.
4. So. f(0) = -1 and f(10) = 109, so the sum is 1/2(-1 - 1/109) = 1/2 × (-110/109) = -55/109, which is option (b).

The idea to keep. When the denominator is a quartic like r⁴ + ar² + 1, make it a difference of squares. If the two factors differ by a multiple of the numerator, the series telescopes.

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