I am a little baffled by this formula
1+....+n = [n(n+1)]/2
I tried to figure it out, but I am left nowhere comprehensible. Would someone be so kind to explain how the formula came about without induction?

- nincompoop

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- katieb

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- austinL

Um, wow.... I haven't the fuzziest. If I may ask, what course is this for?

- nincompoop

"Intro to Mathematical analysis" in spivak's calculus text @austinL

- nincompoop

honest to truth openstudy reintroduced me to my old interest in math

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## More answers

- KingGeorge

Pair the elements up in the following way.
\[1+2+...+n=(1+n)+(2+(n-1))+...x\]Where \(x\) is either \(\frac{n+1}{2}\) if \(n\) is odd, and \((\frac{n}{2}-1)+(\frac{n}{2}+1)=n\) if \(n\) is even.
Each term (except \(x\)) is equal to \(n+1\), and there are \(n/2\) terms if \(n\) is even, immediately giving us the formula\[\frac{n(n+1)}{2}.\]If \(n\) is odd, then we have \(\frac{n-1}{2}\) terms, giving us\[\frac{(n+1)(n-1)}{2}+\frac{n+1}{2}=\frac{n^2+n}{2}=\frac{n(n+1)}{2}\]

- KingGeorge

I hope that was clear enough :/

- nincompoop

Thanks, KG. I'll review what you typed later when I get home. The Safari browser in my iPhone is rendering the latex with errors

- KingGeorge

Well the premise can be done without latex. Basically divide the sum into pairs in the following way
1+2+...+n=(1+n)+(2+(n-1))+(3+(n-2))+...
Then you just have to consider the cases when n is even or odd.

- nincompoop

|dw:1378868007360:dw|

- nincompoop

so our formula is the sum of the sequence up to the whatever nth value?

- nincompoop

|dw:1378868542970:dw|

- nincompoop

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427enudproao7e
with that in mind, let me understand what you typed, KG

- nincompoop

|dw:1378869063985:dw|

- nincompoop

@KingGeorge

- nincompoop

@jim_thompson5910 or @satellite73 or @wio

- nincompoop

my earlier attempt was to figure out if the formula somewhat reverts back to n+1, but I wound up with something else. With KG's help, did I somehow do it correctly?

- nincompoop

my earlier attempt was to figure out if the formula somewhat reverts back to n+1, but I wound up with something else. With KG's help, did I somehow do it correctly? Because I haven't got the slightest clue if I am even understanding things correctly.

- anonymous

ok this is the formula for adding successive whole numbers starting at 1
what exactly are you supposed to do? prove it? use it?

- nincompoop

prove the formula, but I was attempting to do it without using the induction method indicated in the book.

- anonymous

without induction it is pretty obvious
there are a couple ways to do it

- anonymous

\[\sum_{k=1}^nk=1+2+3+4+...+(n-2) + (n-1)+n\]\[\sum_{k=1}^nk=n+(n-1)+(n-2)+...+3+2+1\]
\[2\sum_{k=1}^n=\overbrace{(n+1)+(n+1)+(n+1)+...+(n+1)+(n+1)+(n+1)}^{\text{n summands}}\]
\[2\sum_{k=1}^nk=n(n+1)\]

- anonymous

that is one way
there is another

- nincompoop

What is the other way?
in the page 22 in the 2nd chapter of the prologue of Spivak's calculus in the 4th edition, he stated that the method by which the formula was discovered remains a mystery. http://finedrafts.com/files/math/calculus/Spivak/Michael%20Spivak%20-%20Calculus.pdf

- anonymous

ok here is another way
it is actually a way to find all summation formulas of this type

- anonymous

start one step up
\[\sum_{k=1}^n(k^2-(k-1)^2)\]

- anonymous

after a moments thought, you see that this is a telescoping sum and that it is pretty clear that \[\sum_{k=1}^n(k^2-(k-1)^2)=n^2\]

- anonymous

let me know if that is clear or not, then we can finish it

- nincompoop

it is clear

- anonymous

ok now we do a tiny bit of algebra, that has nothing to do with sums
\[k^2-(k-1)^2=k^2-(k^2-2k+1)=2k-1\]

- anonymous

this tells us that
\[\sum_{k=1}^n(2k-1)=n^2\] also a well known fact

- nincompoop

see that is what I was getting earlier, but I couldn't quite understand it

- anonymous

which step?

- nincompoop

the last one

- nincompoop

second to the last, sorry

- anonymous

the algebra?

- nincompoop

yes

- anonymous

hmm i wrote it out
all i did was take \((k-1)^2\), expand it as \((k-1)(k-1)=k^2-2k+1\) and then subtract the result from \(k^2\)

- anonymous

\[k^2-(k^2-2k+1)=k^2-k^2+2k-1=2k-1\] that is algebra only

- anonymous

clear or no?

- nincompoop

clear. I am looking at my notes

- anonymous

ok so now we have
fact 1 \(\sum k^2-(k-1)^2=n^2\) and
fact 2) \(k^2-(k-1)^2=2k-1\) putting them together gives
\[\sum_{k=1}^n2k-1=n^2\] or
\[2\sum_{k=1}^nk-\sum_{k=1}^n1=n^2\]

- anonymous

good or no?

- nincompoop

good

- anonymous

ok
last fact
\[\sum_{k=1}^n1=n\]

- anonymous

now we have
\[2\sum_{k=1}^nk-\sum_{k=1}^n1=n^2\] is
\[2\sum_{k=1}^nk-n=n^2\]and we can solve this equation for \(\sum k\) using elementary algebra

- anonymous

add \(n\) and divide by \(2\) and get
\[\sum_{k=1}^nk=\frac{n^2+n}{2}\] finished

- anonymous

now suppose i want
\[\sum k^2\]
repeat exactly this process
start one step up with
\[\sum_{k=1}^nk^3-(k-1)^3=n^3\] expand etc

- anonymous

i will leave that as a homework exercise

- nincompoop

Thank you so much. I am going to review and digest this for a while.

- anonymous

you can keep going using this technique to find \(\sum k^3\) etc etc

- anonymous

yw

- nincompoop

I want to be able to redo this in the same manner you did. I hope I can save all of it in PDF

- anonymous

##### 1 Attachment

- anonymous

9 - 12

- anonymous

also the last page

- nincompoop

reviewing them.

- nincompoop

I couldn't thank you enough

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