0

I don't even know how to begin this question as I've never had any mathematical education higher than some rudiment differentials but it comes down to: what should I look for? I was trying to find a formula for number of orbitals given the principal quantum number. It's just a sum of consecutive odd numbers. After writing some examples I noticed it's just $n^2$ but I wasn't able to prove it. I tried writing it down as: $$ \sum^n_{k=1}(2k-1)=n^2 $$ And I can see that's true but why? Why is a sum of $n$ consecutive odd numbers equal to $n^2$? Where do I even start proving it?

Jakub Jędrusiak
  • 133
  • 1
    See https://math.stackexchange.com/a/639079/386635 – Kolja Sep 18 '21 at 20:25
  • https://math.stackexchange.com/q/136237/42969 – Martin R Sep 18 '21 at 20:26
  • 1
    I voted to close this because it is a duplicate. In fact, the link provided by @Kolja was also closed as a duplicate but I also recommend that link because it has a nice simple geometric proof that will enable you to see why immediately. – John Douma Sep 18 '21 at 20:28
  • Thank you all, I've made it. I hadn't realised I can split $\sum^n_{k=1}(2k-1)$ into $2 \times \sum^n_{k=1}k + \sum^n_{k=1}1$ and it made everything much simpler, noticing an arithmetic progression etc. I'm still adjusting to sums being written with sigmas so I'm not really fluent with it. Nevertheless it was quite fun and enlightening, thank you. – Jakub Jędrusiak Sep 18 '21 at 20:52
  • 1
    Alternative : think telescoping. $1 = 1^2 - 0^2.3 = 2^2 - 1^2.5 = 3^2 - 2^2~~\cdots.$ – user2661923 Sep 18 '21 at 21:26

0 Answers0