# How do summations work?

##### 1 Answer

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#oo# , 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)^(oo) 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)^(oo) x^n + sum_(n=2)^(oo) (2x)^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)^(oo) x^n + sum_(n=2)^(oo) (2x)^n#

#= 1 + x + sum_(n=2)^(oo) [x^n + (2x)^n]#

Lastly, a more explicit example would be:

#color(blue)(sum_(n=0)^(5) (2^n + 2n))#

#= [2^0 + 2(0)] + [2^1 + 2(1)] + [2^2 + 2(2)] + [2^3 + 2(3)] + [2^4 + 2(4)] + [2^5 + 2(5)]#

#= sum_(n=0)^(5) 2^n + sum_(n=0)^(5) 2n#

#= [2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5] + [2(0) + 2(1) + 2(2) + 2(3) + 2(4) + 2(5)]#

#= [1 + 2 + 4 + 8 + 16 + 32] + [0 + 2 + 4 + 6 + 8 + 10]#

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

#= color(blue)(93)#