(2³-1³)/(1 × 7)+(4³-3³+2³-1³)/(2 × 11)+(6³-5³+4³-3³+2³-1³)/(3 × 15)+…+ (30³-29³+28³-27³+…+2³-1³)/(15 × 63) is equal to _____________.

Solution

Maths sir · 4 steps

Find the r-th numerator in closed form; it contains r(4r+3), so each term collapses to just r.

1. General term. The r-th numerator is Σₖ₌₁ʳ[(2k)³-(2k-1)³]. The denominators are 1 × 7, 2 × 11, 3 × 15, so the r-th one is r(4r+3). The last term has r = 15, since 4 × 15 + 3 = 63.
2. Expand one pair. (2k)³-(2k-1)³ = 8k³-(8k³-12k²+6k-1) = 12k²-6k+1. Sum this from k=1 to r using Σ k² and Σ k.

   2r(r+1)(2r+1)-3r(r+1)+r = r(4r²+3r) = r²(4r+3)

3. Cancel. The r-th term is r²(4r+3)/r(4r+3) = r. Check with r=1: (8-1)/7 = 1.
4. So. The sum is 1+2+…+15 = (15 × 16)/2 = 120.

The idea to keep. When a series looks frightening, write its r-th term in closed form first. Here the numerator factorises to match the denominator, and the whole series becomes Σ r.

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