nincompoop
  • nincompoop
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
austinL
  • austinL
Um, wow.... I haven't the fuzziest. If I may ask, what course is this for?
nincompoop
  • nincompoop
"Intro to Mathematical analysis" in spivak's calculus text @austinL
nincompoop
  • nincompoop
honest to truth openstudy reintroduced me to my old interest in math

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

KingGeorge
  • 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
  • KingGeorge
I hope that was clear enough :/
nincompoop
  • 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
  • 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
  • nincompoop
|dw:1378868007360:dw|
nincompoop
  • nincompoop
so our formula is the sum of the sequence up to the whatever nth value?
nincompoop
  • nincompoop
|dw:1378868542970:dw|
nincompoop
  • nincompoop
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427enudproao7e with that in mind, let me understand what you typed, KG
nincompoop
  • nincompoop
|dw:1378869063985:dw|
nincompoop
  • nincompoop
@KingGeorge
nincompoop
  • nincompoop
@jim_thompson5910 or @satellite73 or @wio
nincompoop
  • 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
  • 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
  • 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
  • nincompoop
prove the formula, but I was attempting to do it without using the induction method indicated in the book.
anonymous
  • anonymous
without induction it is pretty obvious there are a couple ways to do it
anonymous
  • 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
  • anonymous
that is one way there is another
nincompoop
  • 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
  • anonymous
ok here is another way it is actually a way to find all summation formulas of this type
anonymous
  • anonymous
start one step up \[\sum_{k=1}^n(k^2-(k-1)^2)\]
anonymous
  • 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
  • anonymous
let me know if that is clear or not, then we can finish it
nincompoop
  • nincompoop
it is clear
anonymous
  • 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
  • anonymous
this tells us that \[\sum_{k=1}^n(2k-1)=n^2\] also a well known fact
nincompoop
  • nincompoop
see that is what I was getting earlier, but I couldn't quite understand it
anonymous
  • anonymous
which step?
nincompoop
  • nincompoop
the last one
nincompoop
  • nincompoop
second to the last, sorry
anonymous
  • anonymous
the algebra?
nincompoop
  • nincompoop
yes
anonymous
  • 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
  • anonymous
\[k^2-(k^2-2k+1)=k^2-k^2+2k-1=2k-1\] that is algebra only
anonymous
  • anonymous
clear or no?
nincompoop
  • nincompoop
clear. I am looking at my notes
anonymous
  • 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
  • anonymous
good or no?
nincompoop
  • nincompoop
good
anonymous
  • anonymous
ok last fact \[\sum_{k=1}^n1=n\]
anonymous
  • 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
  • anonymous
add \(n\) and divide by \(2\) and get \[\sum_{k=1}^nk=\frac{n^2+n}{2}\] finished
anonymous
  • 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
  • anonymous
i will leave that as a homework exercise
nincompoop
  • nincompoop
Thank you so much. I am going to review and digest this for a while.
anonymous
  • anonymous
you can keep going using this technique to find \(\sum k^3\) etc etc
anonymous
  • anonymous
yw
nincompoop
  • 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
  • anonymous
anonymous
  • anonymous
9 - 12
anonymous
  • anonymous
also the last page
nincompoop
  • nincompoop
reviewing them.
nincompoop
  • nincompoop
I couldn't thank you enough

Looking for something else?

Not the answer you are looking for? Search for more explanations.