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:
$2S = n(2n+2)$
$S = n(n+1)$.

$$1+3+5+\cdots+(2n+1) = \sum_{k=0}^n (2 k + 1) = 2 \binom{n+1}{2} + (n+1) = (n+1)^2$$
