Find a formula for $1 + 3 + 5 + ... +(2n - 1)$, for $n \ge 1$, and prove that your formula is correct.

Question

I think the formula is $n^2$.

Define $p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$

Then $p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n = (n + 1)^2$

So $p(n + 1): n^2 + 2n = (n + 1)^2$

The equality above is incorrect, so either my formula is wrong or my proof of the implication is wrong or both.

Can you elaborate?

Thanks.

Answer 1

The issue here is that $p(n+1)$ is note the statement $$ 1+3+5+\cdots+(2n-1)+2n=(n+1)^2; $$ it is the statement $$ 1+3+5+\cdots+(2n-1)+(2n+1)=(n+1)^2. $$

Why? The left side of your formula is the sum of all odd numbers between $1$ and $2n-1$. So, when you replace $n$ by $n+1$, you get the sum of all odd numbers between $1$ and $2(n+1)-1=2n+1$.

Answer 2

$p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2$

$p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n+1 =n^2+2n+1= (n + 1)^2$ next term after $2n-1$ is $2n+1$ is not $2n$ as you mean

Answer 3

Rewrite the series as $$\sum\limits_{i=1}^n 2i-1$$ Continuing, $$...=2\sum\limits_{i=1}^n i - \sum\limits_{i=1}^n 1$$ $$=2\frac{n(n+1)}{2} - n$$ $$=n^2+n-n$$ $$=n^2$$

As you expected.

Answer 4

$n^2 - (n-1)^2 = 2n - 1$ might be the main part of a proof.