How is sigma notation used to find the value of a series?

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How is sigma notation used to find the value of a series?

Mathematics
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Sigma notation denotes a sum of a series.
This "sigma notation" in general works this way: \(\Large\color{black}{ \displaystyle \sum_{ n=\color{orangered}{ \rm k} }^{ \color{orangered}{ \rm z} } ~ f(n)=a_\color{orangered}{ \rm k}+a_{\color{orangered}{ \rm k}+1}+a_{\color{orangered}{ \rm k}+3}+~...~+a_{\color{orangered}{ \rm z}-1}+a_{\color{orangered}{ \rm z}}}\)
should I give a particular example, or examples?

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Or do you have an example that you are working on already?
No, I just need to know how it used to find the value of a series
I will use a more easy explanation then.
\(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{i} }^{ \color{blue}{w} } ~ f(n)}\) f(n) is any pattern that involves n (e.g. adding a number every time, multiplying times a number every time, and other patterns) i is the number of the term from which you start. w is the number of term till which (and including which) you are adding. ------------------------------------------------------- For example: \(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{1} }^{ \color{blue}{10} } 5n= \large (5\cdot \color{red}{1})+(5\cdot \color{red}{2})+(5\cdot \color{red}{3})+(5\cdot \color{red}{4}) +(5\cdot \color{red}{5})+\\ \large ~~~~~~~~~~~~~~~~~~~~~~~~~(5\cdot \color{red}{6})+(5\cdot \color{red}{7})+(5\cdot \color{red}{8})+(5\cdot \color{red}{9})+(5\cdot \color{red}{10})}\)
and then calculate that (I am demostrating another helpful technique that is not THTA relevant now, but helpful) (5⋅1)+(5⋅2)+(5⋅3)+(5⋅4)+(5⋅5)+ (5⋅6)+(5⋅7)+(5⋅8)+(5⋅9)+(5⋅10)= (1+2+3+4+5+6+7+8+9+10)•5= this step uses the formula of a sum for arithmetic sequence first term, 1, plus last term, 10, divided by 2 - which is altogther the average term (this avergae term is in gray), and then times the number of term (this should follow logically too: average•total number of terms=total value) [ `(10+1)•½`•10 ]•5= [ (10+1)•5 ]•5= [ 11•5 ]•5= 11 • 25= 250+25= \(\LARGE 275\)
Another example, of a sigma notation meaning is: \(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{3} }^{ \color{blue}{6} } 5^n= \large (5^\color{red}{3})+(5^\color{red}{4}) +(5^\color{red}{5})+(5^\color{red}{6})+(5^\color{red}{7})}\) Note that I am starting from 5³, not from 5¹, because n=3 on the bottom tells me to do so.
Oh, the blue number on top should be 7, it is a TYPO
So we added till and including the 7th term, AND started adding from the 3rd term/
the point is that you can treat it as a linear operator, i.e. it distributes over linear combinations of pairs of sequences \(x_n,y_n\) $$\sum (Ax_n+By_n)=A\sum x_n+\sum y_n$$so if we have a complicated series like, say, \(a_n=3n^2+4n+6\), then, say, $$\sum_{n=1}^{10} (3n^2+4n+6)=3\left(\sum_{n=1}^{10} n^2\right)+4\left(\sum_{n=1}^{10} n\right)+6\left(\sum_{n=1}^{10} 1\right)$$
oops, the first line should read $$\sum (Ax_n+By_n)=A\sum x_n+B\sum y_n$$
and then this complex sequence \(a_n=3n^2+4n+6\) can be summed for the series simply by knowing how to sum its simple constituent parts, \(n^2,n,1\): $$\sum_{n=1}^{10} n^2=\frac{10(11)(21)}6=385\\\sum_{n=1}^{10}n=\frac{10(11)}2=55\\\sum_{n=1}^{10}1=10$$
so $$\sum_{n=1}^{10}(3n^2+6n+4)=3\cdot385+6\cdot55+4\cdot10=1525$$ ... and I think we can all agree that the sums of \(n^2,n,1\) are way easier than the sum of \(3n^2+4n+6\), even when done by hand without knowing the identities I used above
this is the power of the notation -- it lets us see how the distributive and commutative properties of addition can turn complicated sums into more simple ones in a very nice, neat manner, without actually having to write out all the terms as in a traditional sum (which is useful for when this is not possible, e.g. in infinite sums)
Thank you guys so much!!

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