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:
$2S = n(2n+2)$
$S = n(n+1)$.
 A: 
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: The last of $n$ numbers starting from $1, 3, 5, \ldots$ is $2n-1$, not $2n+1$.  Replace accordingly and you should obtain the right answer.
A: $$S=1+3+5+7+\cdots 2r+1$$
$$S=2r+1 +2r-1+2r-3+2r-5+2r-7 +\cdots 1  $$
$$2S=2(2r+1)+r(2r)$$
$$S=r^2+2r+1=(r+1)^2$$
Instead of $r^{th}$ term  you have written $(r+1)^{th}$ term.
This can also be verified using 
$$a_r=a + (n-1)d$$
$$a_r=2r-1 \qquad a_{r+1}=2r +1 $$
You can do it easily by this
We got genral term as $(2r-1)$
$$\sum_{r=1}^n2r-1$$
$$2\sum_{r=1}^n r-\sum_{r=1}^n1$$
A: Only have $n-1$ numbers of (2n+2)
therefore:
2S = (n-1)(2n+2)
S = (n-1)(n+1)
Reason:
when $n=1$, then first term is 3 rather than 1.
Therefore, you only have $n-1$ numbers of (2n+2).
A: It is very simple, the sequence 1,3,5,........(2n+1) is arithmetic sequence
having first term let say f = 1, last term say l = (2n+1) and common difference d = 2 , {(3-1),(5-3)......so on}  and the sequence have n+1 terms so it can be represented as 1,3,5,..........(2n-1),(2(n+1)-1)
we can use formula for sum of arithmetic sequence having first term(f), last term(l) and total number of terms(n) as:
sum = n(f+l)/2
so,
 sum = (n+1)(1+2n+1)/2
      = (n+1)(n+1)
      = (n+1)^2

Hope it helps.......
A: $$1+3+5+\cdots+(2n+1) = \sum_{k=0}^n (2 k + 1) = 2 \binom{n+1}{2} + (n+1) = (n+1)^2$$
A: $$S=\sum^n_{k=1} (2n-1) \implies \sum^n_{k=1} 2n- \sum^n_{k=1} 1 \implies {2(n)(n+1)\over2} - n \implies n^2+n-n\implies n^2$$
