The sum Σₖ₌₁²⁰ ( 1 + 2 + 3 + ... + k ) is :

Solution

Maths sir · 4 steps

Each bracket is a triangular number, so write it as k(k+1)/2 and then sum two standard series.

1. Close the bracket. The bracket is the sum of the first k natural numbers, so 1 + 2 + … + k = k(k+1)/2. The sum becomes Σₖ₌₁²⁰ (k² + k)/2.

   Σₖ₌₁²⁰ k(k+1)/2 = 1/2(Σₖ₌₁²⁰ k² + Σₖ₌₁²⁰ k)

2. Standard sums. With n = 20: Σ k² = n(n+1)(2n+1)/6 = (20 × 21 × 41)/6 = 2870 and Σ k = n(n+1)/2 = (20 × 21)/2 = 210.
3. Check by formula. The two pieces combine into n(n+1)(n+2)/6. For n = 20 this is (20 × 21 × 22)/6 = 9240/6 = 1540.
4. So. The sum is 1/2(2870 + 210) = 3080/2 = 1540.

The common slip: 3080. Dropping the 1/2 from k(k+1)/2 gives 2870 + 210 = 3080, which is double the answer.

The idea to keep. A sum of sums: close the inner sum into a formula in k first, then use Σ k and Σ k². Adding triangular numbers always gives n(n+1)(n+2)/6.

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