Question

How do you represent the sum of first \(n\) odd numbers through sigma notation?

Answer 1

I suppose it is something related to \(2n + 1\) where \(n\) starts at 0?

Answer 2

\[
\sum_{i=1}^n 2i-1
\]

Answer 3

I see.

Answer 4

So:\[ \color{Black}{\Rightarrow \sum_{i = 1}^{n} 2i - 1} = n^2\]

Answer 5

If it had been \(2i + 1\), then \(i\) would have started at \(0\) and finished at \(n - 1\) right?

Answer 6

Or you can also take as:

\[\large \sum_{k=0}^n 2k+1\]

Answer 7

Oh wait. I just got confused.

Answer 8

Yes, if you start at 0 the sum should only go to n-1.

Answer 9

We have a conflict of opinions right there.

Answer 10

Let me try it out for \(n = 2\).

Answer 11

\[ \sum_{i = 0}^{2 - 1} 2i + 1 = 2(0) + 1 + 2(1) + 1 = 1 + 3\]

Answer 12

Yay! I was correct!

Answer 13

Thank you @nbouscal :)

Answer 14

Yes there will come (n-1) sorry..