Direct Proof for sum of $n$ integers equation?

Question

I am trying to prove by direct proof that

$$3+5+7+\ldots+(2n+1)=n(n+2)$$

for all natural numbers $n$. I figured out how to do it by induction, but I know it can be done directly and I can't figure out.

Thanks in advance!

Answer 1

Do what Gauss did, reverse the sequence and add it with the original. Now each term is 2n + 4, and there are n such terms. The sum is n(2n+4). Since this results from two sequences, divide by 2 and you get the answer.

Answer 2

Note that $$\sum\limits_{k=1}^n (2k+1)=2\sum\limits_{k=1}^n k+\sum\limits_{k=1}^n1=2\frac{n(n+1)}{2}+n=n(n+2)$$

where we have used the fact that $\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$.

Answer 3

Here is an elegant "picture proof" that $\sum_\limits{i=1}^n (2i-1) = n^2$. So you have, $\sum_\limits{i=1}^n (2i+1) = \sum_\limits{i=1}^n (2i-1) + (2n+1) - 1 = n^2 + 2n$

Answer 4

$$\begin{align}1+3+5+7+\cdots+(2n-1)&=\sum_{r=1}^n (2r-1)=\sum_{r=1}^n r^2-(r-1)^2=n^2\\ 3+5+7+\cdots+(\color{red}{2n}+\color{blue}1)&=\underbrace{\color{blue}1+3+5+7+\cdots+(2n-1)}_{n^2}+(\color{red}{2n})\\&=n^2+2n\\&=n(n+2)\qquad\blacksquare\end{align}$$