The sum 1²-2 · 3²+3 · 5²-4 · 7²+5 · 9²-…+15 · 29² is _________.

Solution

Maths sir · 4 steps

Signs alternate, so pair each odd term with the next even term; the 15th term is left alone.

1. General term. The nth term is (-1)ⁿ⁺¹ n (2n-1)² for n = 1 to 15. There are 15 terms, an odd count, so 7 pairs form and the last term 15 · 29² stays unpaired.
2. Simplify one pair. Take n = 2k-1 (positive) and n = 2k (negative), for k = 1 to 7. Expanding, the k³ terms cancel. Check with k=1: 1 - 2 · 9 = -17 = -48 + 40 - 9.

   (2k-1)(4k-3)² - 2k(4k-1)² = -48k² + 40k - 9

3. Sum the pairs. With Σ k² = 140 and Σ k = 28 for k = 1 to 7: -48(140) + 40(28) - 9(7) = -6720 + 1120 - 63 = -5663.
4. So. Add the unpaired last term 15 · 841 = 12615: the sum is -5663 + 12615 = 6952.

The common slip: -5663. Forgetting the unpaired 15th term gives only the sum of the 7 pairs, which is -5663.

The idea to keep. For an alternating series, pair neighbouring terms so the highest powers cancel. Count the terms first: an odd count leaves one term outside the pairs.

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