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anonymous
 one year ago
How is sigma notation used to find the value of a series?
anonymous
 one year ago
How is sigma notation used to find the value of a series?

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Sigma notation denotes a sum of a series.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1This "sigma notation" in general works this way: \(\Large\color{black}{ \displaystyle \sum_{ n=\color{orangered}{ \rm k} }^{ \color{orangered}{ \rm z} } ~ f(n)=a_\color{orangered}{ \rm k}+a_{\color{orangered}{ \rm k}+1}+a_{\color{orangered}{ \rm k}+3}+~...~+a_{\color{orangered}{ \rm z}1}+a_{\color{orangered}{ \rm z}}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1should I give a particular example, or examples?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Or do you have an example that you are working on already?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I just need to know how it used to find the value of a series

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I will use a more easy explanation then.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{i} }^{ \color{blue}{w} } ~ f(n)}\) 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: \(\Large\color{black}{ \displaystyle \sum_{ n=\color{red}{1} }^{ \color{blue}{10} } 5n= \large (5\cdot \color{red}{1})+(5\cdot \color{red}{2})+(5\cdot \color{red}{3})+(5\cdot \color{red}{4}) +(5\cdot \color{red}{5})+\\ \large ~~~~~~~~~~~~~~~~~~~~~~~~~(5\cdot \color{red}{6})+(5\cdot \color{red}{7})+(5\cdot \color{red}{8})+(5\cdot \color{red}{9})+(5\cdot \color{red}{10})}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and 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= \(\LARGE 275\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Another 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.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Oh, the blue number on top should be 7, it is a TYPO

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So we added till and including the 7th term, AND started adding from the 3rd term/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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, $$\sum_{n=1}^{10} (3n^2+4n+6)=3\left(\sum_{n=1}^{10} n^2\right)+4\left(\sum_{n=1}^{10} n\right)+6\left(\sum_{n=1}^{10} 1\right)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, the first line should read $$\sum (Ax_n+By_n)=A\sum x_n+B\sum y_n$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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, \(n^2,n,1\): $$\sum_{n=1}^{10} n^2=\frac{10(11)(21)}6=385\\\sum_{n=1}^{10}n=\frac{10(11)}2=55\\\sum_{n=1}^{10}1=10$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$\sum_{n=1}^{10}(3n^2+6n+4)=3\cdot385+6\cdot55+4\cdot10=1525$$ ... 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you guys so much!!
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