# How do summations work?

Jun 17, 2017

I think you're asking how the notation works, but I will also cover some examples of adding with it.

The basic structure of it is:

sum_(n=a)^(N) f(x_1, x_2, . . ., x_m ; n)

• The $n$, or whatever letter you want, is an index, and it indicates what spot we are in for the sum. What $n$ is written to be equal to is the first index in the sum (whatever $a$ is).
• The $N$ in general indicates the upper limit to the sum. The sum could be written to go towards $\infty$, or it could stop at the finite number $N$, depending on context.
• f(x_1, x_2, . . . , x_m; n) is a function of some number of variables indicated by the ${x}_{i}$ ($i = 1 , 2 , . . . , m$), and usually is also a function of the index $n$.

For example, if we write

${\sum}_{n = 0}^{\infty} {x}^{n}$,

we are writing a condensed form of:

${x}^{0} + {x}^{1} + {x}^{2} + \cdot \cdot \cdot$

$= 1 + x + {x}^{2} + \cdot \cdot \cdot$

Or, if we combine two sums, like writing

${\sum}_{n = 0}^{\infty} {x}^{n} + {\sum}_{n = 2}^{\infty} {\left(2 x\right)}^{n}$,

sums are commutative and their indices can be linked together, so this can also be written as:

$= {\sum}_{n = 0}^{1} {x}^{n} + {\sum}_{n = 2}^{\infty} {x}^{n} + {\sum}_{n = 2}^{\infty} {\left(2 x\right)}^{n}$

$= 1 + x + {\sum}_{n = 2}^{\infty} \left[{x}^{n} + {\left(2 x\right)}^{n}\right]$

Lastly, a more explicit example would be:

$\textcolor{b l u e}{{\sum}_{n = 0}^{5} \left({2}^{n} + 2 n\right)}$

$= \left[{2}^{0} + 2 \left(0\right)\right] + \left[{2}^{1} + 2 \left(1\right)\right] + \left[{2}^{2} + 2 \left(2\right)\right] + \left[{2}^{3} + 2 \left(3\right)\right] + \left[{2}^{4} + 2 \left(4\right)\right] + \left[{2}^{5} + 2 \left(5\right)\right]$

$= {\sum}_{n = 0}^{5} {2}^{n} + {\sum}_{n = 0}^{5} 2 n$

$= \left[{2}^{0} + {2}^{1} + {2}^{2} + {2}^{3} + {2}^{4} + {2}^{5}\right] + \left[2 \left(0\right) + 2 \left(1\right) + 2 \left(2\right) + 2 \left(3\right) + 2 \left(4\right) + 2 \left(5\right)\right]$

$= \left[1 + 2 + 4 + 8 + 16 + 32\right] + \left[0 + 2 + 4 + 6 + 8 + 10\right]$

$= 1 + 2 + 4 + 8 + 16 + 32 + 0 + 2 + 4 + 6 + 8 + 10$

$= \textcolor{b l u e}{93}$