How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$. 
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
 A: The second equality can be understood by letting the number of terms in the second sum be $m$. Here $m=2n$.
$n$ is just a symbol. A sum can be represented in different ways and $n$ can mean different things. In the first case, it means number of terms. In the second second case, it means number of terms divided by $2$. 
Hence if $n=4$, $$1+2+ \ldots + n = 1+2+ \ldots + 4=\frac{n(n+1)}2=10$$
If $n=2$,
$$1+2+ \ldots + 2n = 1+2+ \ldots + 4=\frac{(2n)(2n+1)}2=10$$
A: Compare:
$$\color{red}{1+2+\cdots +n}=\frac{n(n+1)}{2}$$
and
$$\color{red}{1+2+\cdots+n}+\color{blue}{(n+1)+(n+2)+\cdots+2n}=$$
$$\color{red}{1+2+\cdots+n}+\color{blue}{(1+2+\cdots+n)+n\cdot n}=$$
$$\color{red}{\frac{n(n+1)}{2}}+\color{blue}{\frac{n(n+1)}{2}+n^2}=\frac{2n(2n+1)}{2}$$
A: The formula $1+2+3+...+n$ mean the "sum of all integers starting with $1$ and going up to $n$". You can substitute any positive integer you want in place of $n$, for instance by substituting $n=6$ you get $1 + 2 + 3 + ... + 6$, which ordinarily we would write out more explicitly as $1+2+3+4+5+6$, and them sum to get $21$.
Similarly, the formula $1+2+3+....+BLAH$ to represent the "sum of all integers starting with $1$ and going up to BLAH. You can substitute any positive integer you want in place of BLAH.
Finally the formula $1+2+3+...+2n$ mean "the sum of all integers starting with $1$ and going up to $2n$". You can substitute any positive integer you want in place of $n$, and then do the arithmetic to get $2n$, and then substitute that amount. For instance, if you take $n=3$ and substitute that into the expression $2n$ then you get $2n = 2 \cdot 3 = 6$, and then when you substitute $2n=6$ into the formula $1+2+3+...+2n$ you get $1+2+3+...+6 = 21$.
A: Let $n = 4$ and $m = 2n = 8$.
Then $1 + 2 + 3 + 4 + 5  +6+7+8 = 36$.
$1+2+3+4+5+6+7+8 = \frac {8(8+1)}{2}= \frac {2*4(2*4 + 1)}2 = 36$.
$1 + 2 + 3 + ....... + m = \frac {m(m+1)}{2}$ and if $m = 8$ then $1 + 2 + 3 +....... + 8 = \frac {8(8+1)}{2} = 36$.
And $1+2+3+...... + 2n -1 + 2n = \frac {2n(2n+1)}2$ and if $n= 4$ then $1 + 2+ 3 + ...... + 8 = \frac {(2*4)(2*4 + 1) } 2 = 36$.
It's nothing profound or difficult.
$1 + 2 + 3 +.......... + whatever = \frac {whatever(whatever + 1)}2$.
And $whatever$ can be .... whatever you want.
