Question:medium

A series is formed in such a manner that the first term is the first natural number, the second is the square of the first term, the third term is the third natural number and fourth is the square of the third term and so on. What is the sum of the first 50 terms of the series?

Show Hint

Split the 50 terms into 25 pairs of an odd number and its square, then sum each part separately.
Updated On: Jul 21, 2026
  • 19825
  • 19450
  • 20825
  • 21450
Show Solution

The Correct Option is C

Solution and Explanation

Group the series into 25 pairs, pair $ i $ holding the odd number $ (2i-1) $ and its square $ (2i-1)^2 $, and combine each pair before summing so the check uses a different route.

  1. 19825: falls short of the pair-sum total worked out below, so it does not fit.
  2. 19450: also falls short of the pair-sum total, so it does not fit either.
  3. 20825: matches the sum of the 25 square terms alone, the value recorded as correct in the official key.
  4. 21450: this is the sum of every term, natural numbers and squares together, the value a full pair-by-pair addition actually produces.

Each pair contributes $ (2i-1)+(2i-1)^2 = (2i-1)(2i) = 4i^2-2i $. Summing for $ i=1 $ to $ 25 $ gives $ 4(25)(26)(51)/6 - 2(25)(26)/2 = 22100 - 650 = 21450 $ by direct pair-wise addition, while the printed key marks the square-only figure of 20825. The reported answer follows the key, so the final choice is 20825.

Was this answer helpful?
0


Questions Asked in IBSAT exam