The sum of the first n odd natural numbers The sum of the first n odd natural numbers: $1+3+5+7+...+(2n-1)$.
My question is: Why do we write the last element as $2n-1$ ? Why don't we write $1+3+5+7+...+(2n+1)$ ?
 A: That is a simple matter of counting. The first natural odd number is
$$2\cdot(1) - 1 = 1$$
The second is
$$2\cdot(2) - 1 = 3$$
The third is
$$2\cdot(3) - 1 = 5$$
...
The 57th is
$$2\cdot(57) - 1 = 113$$
And the $n $th is
$$2\cdot(n) - 1 = 2n - 1$$
So the first $n $ odd natural numbers end with  $(2n-1) $.
A: We write $2n-1$ as we are using $n$ terms in the expression. The 1st term is 1, the 2nd term is 3, the 3rd term is 5, and so on until the nth term is $2n-1$.
This can be seen by looking at the numbers and thinking about what formula gives that pattern. E.g.
$1\rightarrow1$
$2\rightarrow3$
$3\rightarrow5$
$\cdots$
$n\rightarrow?$
A: I'ts for counting set $\mathbb{N}$ that is used for indexing terms of sequence or series. The $n$-th term of a sequence is $a_n$ and first term generally is began with index $1$, the first member of $\mathbb{N}$.
A: In the expression $2n-1$, the value goes from $1 \to n$ for the values for $n$ to get $1,3,5 ... 2n-1$ 
Since $n$ is defined over $\mathbb N$.
In the expression $2n+1$, if you want to keep the values as in the previous expression you would need to go from $0$ to $n-1$, for the values of $n$.
But that's not how it works by its definition.
Since $n$ is defined over $\mathbb N$, the second expression gives $3,5,7...2n+1$

These two are not the same since $n\in \mathbb N$ always follows natural values from $1$ to $n$; $(1,2,3,...n)$
The second expression can be still used instead of the first one, but only if you say beforehand that $n\in \mathbb W$ and that the value $n$ itself is excluded, leaving the $n-1$ the last value.
But it really makes no sense to do that. You would only make things inconvenient.

W stands for whole numbers, and N stands for natural numbers.
$\mathbb N = \{1,2,3... \}$
$\mathbb W = \{0,1,2,3... \}$
