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How to do summation of k^2 + 1?
How would I get the formula to find the summation of (k+1) for k=0...n? (So it would be 1 + 3 + 5...n^2 + 1)?
would it be n^3/3 + n^2/2 + n/6 + (n+1)?
1 Answer
- alwbsokLv 710 years ago
The sum of k^2 + 1 would be:
1 + 2 + 5 + 10 + 17 + 26 + ...
It would also be the sum of k^2 plus the sum of 1. The sum of k^2 from k = 0 to n is:
(1/6)n(n + 1)(2n + 1)
The sum of 1 from k = 0 to n is:
n + 1
So the answer is:
(1/6)n(n + 1)(2n + 1) + n + 1
= (1/6)(n + 1)(2n^2 + n) + (1/6)(n + 1)(6)
= (1/6)(n + 1)(2n^2 + n + 6)
That's the way I'd personally present the answer. You can also expand:
(1/6)(2n^3 + 3n^2 + 7n + 6)
= (1/3)n^3 + (1/2)n^2 + (7/6)n + 1
which agrees with what you wrote.
