Calculate the value of the sum $1+3+5+\cdots+(2n+1)$ I have been thinking about this for a long time, may I know which step of my thinking is wrong as I do not seems to get the correct answer. If I am not going towards the right direction, may I get some help thanks!

My attempt:

Let $S = 1+3+5+\dotsb+(2n+1)\label{a}\tag{1}.$

Then I rearrange S from the last to first terms:

$S = (2n+1)+(2n-1)+(2n-3)+\dotsb+1\label{b}\tag{2}.$

Adding the two series $(1)+(2)$:

$$2S = (2n+2)+(2n+2)+(2n+2)+\dotsb+(2n+2),$$

I have $n$ copies of $(2n+2)$.

Therefore:

$2S = n(2n+2)$

$S = n(n+1)$.
Build a square in the following manner. On your first step, place $1$ block. On your second step put $3$ blocks around that, and on your third step put $5$ blocks around what you have, and so on. I think it should be pretty easy to see that the sum of the blocks is the area of the square for the step you are on.

In other words, the sum of the first $n$ odd integers is

$$
(2(1) - 1) + (2(2) - 1) + \dotsb + (2(n) - 1) = n^2.
$$

Image source: google "sum of first $n$ odd numbers."
A tip: evaluate the first few partial sums.

$$1, 4, 9, 16, 25, 36, 49\cdots$$

Nothing familiar ?