Question

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

Answer 1

Sigma notation denotes a sum of a series.

Answer 2

This "sigma notation" in general works this way:

\(\Large\color{black}{ \displaystyle \sum_{ n=\color{orangered}{ \rm k} }^{ \color{orangered}{ \rm z} } ~ f(n)=a_\color{orangered}{ \rm k}+a_{\color{orangered}{ \rm k}+1}+a_{\color{orangered}{ \rm k}+3}+~...~+a_{\color{orangered}{ \rm z}-1}+a_{\color{orangered}{ \rm z}}}\)

Answer 3

should I give a particular example, or examples?

Answer 4

Or do you have an example that you are working on already?

Answer 5

No, I just need to know how it used to find the value of a series

Answer 6

I will use a more easy explanation then.

Answer 7

\(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{i} }^{ \color{blue}{w} } ~ f(n)}\)

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.
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For example:
\(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{1} }^{ \color{blue}{10} } 5n= \large (5\cdot \color{red}{1})+(5\cdot \color{red}{2})+(5\cdot \color{red}{3})+(5\cdot \color{red}{4}) +(5\cdot \color{red}{5})+\\ \large ~~~~~~~~~~~~~~~~~~~~~~~~~(5\cdot \color{red}{6})+(5\cdot \color{red}{7})+(5\cdot \color{red}{8})+(5\cdot \color{red}{9})+(5\cdot \color{red}{10})}\)

Answer 8

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= \(\LARGE 275\)

Answer 9

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.

Answer 10

Oh, the blue number on top should be 7, it is a TYPO

Answer 11

So we added till and including the 7th term,
AND
started adding from the 3rd term/

Answer 12

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, $$\sum_{n=1}^{10} (3n^2+4n+6)=3\left(\sum_{n=1}^{10} n^2\right)+4\left(\sum_{n=1}^{10} n\right)+6\left(\sum_{n=1}^{10} 1\right)$$

Answer 13

oops, the first line should read $$\sum (Ax_n+By_n)=A\sum x_n+B\sum y_n$$

Answer 14

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, \(n^2,n,1\): $$\sum_{n=1}^{10} n^2=\frac{10(11)(21)}6=385\\\sum_{n=1}^{10}n=\frac{10(11)}2=55\\\sum_{n=1}^{10}1=10$$

Answer 15

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

Answer 16

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)

Answer 17

Thank you guys so much!!