Question 550732
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I think that you are trying to over think this.  The sigma notation looks scary the first time you see it, but it is really a very simple concept.  It is just a shorthand way to add up a bunch of terms that have a definable pattern.


The sigma itself is nothing more than an upper case Greek letter "S" and it stands for Sum.  But what is to be summed?  Terms that follow the pattern in the expression that is to the right of the *[tex \Large \sum] symbol, that's what.  But how many terms?  That is the meaning of the information above and below the *[tex \Large \sum] symbol.  At the bottom of the symbol is generally an expression that looks something like *[tex \Large i\ =\ 1] which tells you where you are going to start counting your terms, and above the symbol is a number that tells you where you are going to stop counting.  In other words, the number on the bottom is the smallest value that the index variable (the *[tex \Large i] or the *[tex \Large k] typically) can take on, and the top number is the highest or ceiling value that the index variable can be.


Let's look at a couple of examples:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,1}^{42}\,1]


Here the index variable *[tex \Large i] takes on values from 1 to 42 successively but the term definition part DOES NOT have an *[tex \Large i] in in it.  The definition is simply the number 1, so that means that <i>every</i> term is nothing more than a number 1.  So what this means is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,1}^{42}\,1\ =\ 1\ +\ 1\ +\ \cdots\ +\ 1]


And there are 42 terms, and 42 1s add up to -- 42, the answer to life, the universe, and everything.  (Read Doug Adams' <i>Hitchhiker's Guide to the Galaxy</i> if you don't get that)


Next:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{n\,=\,1}^{5}\,n]


Here your index variable is *[tex \Large n] and *[tex \Large n] is also the definition of the terms.  So the first term, when *[tex \Large n] equals 1, is, yes you guessed it, 1.  The second term is 2, and so on until the last term when *[tex \Large n\ =\ 5].  Why stop at 5?  Because that is the end number indicated on top of the symbol.  So let's expand this one:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{n\,=\,1}^{5}\,n\ =\ 1\ +\ 2\ +\ 3\ +\ 4\ +\ 5]


A few seconds looking at that should convince you that the sum is 15.  But let's look at the rule for this sort of series.  Adding up the first 5 integers is pretty simple, but what if you needed to add up the first 100 integers?  I'd rather have a root canal, except that it is so simple it will amaze you.


Consider adding the first 100 positive integers.  You could just go at it with a full on frontal assault, i.e. 1 plus 2 is 3. 3 plus 3 is 6, plus 4 is 10...blah, blah, blah.  Long, boring, and highly prone to error.  Let's think outside the box a little.  1 plus 100 is 101.  2 plus 99 is 101.  3 plus 98 is 101.  Hmmm, 100 numbers, so 50 pairs of numbers, so the sum has to be 101 times 50 = 5050.  Ta Da!!
                           

The general formula for the sum of a series of integers is "the first plus the last times the number of numbers divided by 2."  Let's try this on:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{n\,=\,1}^{5}\,n]


The first number is 1 and the last is 5.  1 plus 5 is 6.  There are 5 numbers, so 6 times 5 is 30.  30 divided by 2 is the expected 15.


That process should enable you to do problems 2 through 4.  For the rest, you need the formula for the sum of consecutive integers squared:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i\,=\,1}^n\,i^2\ =\ \frac{n(n\,+\,1)(2n\,+\,1)}{6}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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