MCQ practice · SSC General Mathematics Board questions · Chapter 13: Finite Series
For the first n natural-
- i. numbers that are odd, the sum is n²
- ii. numbers, the sum of squares is n(n + 1)(2n + 1)/6
- iii. numbers, the sum of cubes is n³(n + 1)³/8
Think it through first, then pick an option.
- a i and ii
- b i and iii
- c ii and iii
- d i, ii and iii
Answer and explanation
Correct answer: a i and ii
Taking only iii as true is wrong: the sum of cubes is {n(n + 1)/2}² = n²(n + 1)²/4, not n³(n + 1)³/8.
Why the other options are wrong
- b i and iii: i and ii are true: odd numbers sum to n² and squares sum to n(n + 1)(2n + 1)/6; iii is false, the cube formula is n²(n + 1)²/4.
- c ii and iii: iii is false: the sum of cubes is n²(n + 1)²/4; and ii is true but left out.
- d i, ii and iii: iii is false: the sum of cubes is not n³(n + 1)³/8, so all three cannot be true.
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