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

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.......
