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Finite series proof

How can ∑r=1nr=12n(n+1)\displaystyle\sum_{r=1}^n r = \frac{1}{2}n(n+1)
be proven, without the usual writing the sequence in reverse and adding up.

Also for ∑r=1nr2\displaystyle\sum_{r=1}^n r^2 and ∑r=1nr3\displaystyle\sum_{r=1}^n r^3.
(edited 11 years ago)
[QUOTE="UKBrah;40850821"]How can
∑r=1nr=1[br]2n(n+1)\displaystyle\sum_{r=1}^n r = \frac{1}[br]{2}n(n+1)
be proven, without the usual writing the
sequence in reverse and adding up.
Also for ∑r=1nr2[/[br]latex]and[latex]∑r=1n[br]r3[/latex].\displaystyle\sum_{r=1}^n r^2[/[br]latex] and [latex]\displaystyle\sum_{r=1}^n[br]r^3[/latex].

Are you asking about proof by induction?
Do you even lift brah?


And see what you make of this sum5.png
Reply 3
Avatar for UKBrah
UKBrah
OP
3
Original post by This Excellency
Are you asking about proof by induction?

I suppose, but for PBI you're given the LHS and the RHS. When you're just given the summation how can you proof that it's equal to 1/2(n)(n+1) without the newfag method? Is it even possible?

Original post by marcus2001
Do you even lift brah?


And see what you make of this sum5.png

revisingatthegym.jpeg

no.

I wrote this without properly reading that is sixthform but I'm posting it anyway just in case might be of use to someone

(edited 11 years ago)
Reply 5
Avatar for 3nTr0pY
3nTr0pY
3
Original post by L'art pour l'art

I wrote this without properly reading that is sixthform but I'm posting it anyway just in case might be of use to someone


Sweet.
Reply 6
Avatar for Keynesian
Keynesian
For just the summation of r, consider a graph with n number of nodes and you wish to calculate the number of arcs that connect all the nodes together. Still with me? Now use the sum of an arithmetic series which is 1/2n(2a+d(n-1)), you know that the first term in essence is n, so a=n and the common difference will be -1 because you are losing a node every time, and the last term stays in terms of n, tidy that up to give the summation formula for r. That is one way of looking at it, no idea how to approach the rest.
Original post by 3nTr0pY
Sweet.
Cheers! :]

Original post by Keynesian
...

Cool... but a question?

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