1^2 + 2^2 + 3^2 + ... + n^2 = ____?
- n(n+1)/2
- n(n+1)(2n+1)/6
- n(n-1)/2
- None of these
Explanation
The sum of squares of first n natural numbers is given by:
1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6
Related MCQs
- 18
- 17
- 16
- None of these
اس سوال کو وضاحت کے ساتھ پڑھیں
- 70
- 69
- 68
- None of these
اس سوال کو وضاحت کے ساتھ پڑھیں
