- #1

PEZenfuego

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So I've been playing with summation a bit and figured out a way to make equations for Ʃ[itex]^{n}_{k=1}[/itex]K and Ʃ[itex]^{n}_{k=1}[/itex]K[itex]^{2}[/itex] That looks odd, so I'll just use Ʃ from now on, but realize that it is always from k=1 to n.

ƩK is a series like 1+2+3+4...+(n-2)+(n-1)+n

and Ʃk[itex]^{2}[/itex] is a series like1[itex]^{2}[/itex]+2[itex]^{2}[/itex]+3[itex]^{2}[/itex]+4[itex]^{2}[/itex]...+(n-2)[itex]^{2}[/itex]+(n-1)[itex]^{2}[/itex]+n[itex]^{2}[/itex]

Anyway, the equation for Ʃk[itex]^{2}[/itex] is (2n[itex]^{3}[/itex]+3n[itex]^{2}[/itex]+n)/6

and the one for ƩK is (n[itex]^{2}[/itex]+n)/2

Now here is what I'm stepping in:

Ʃk[itex]^{2}[/itex]=1[itex]^{2}[/itex]+2[itex]^{2}[/itex]+3[itex]^{2}[/itex]+4[itex]^{2}[/itex]...

So the derivative of this with respect to n would be

dƩk[itex]^{2}[/itex]/dn=2(1)+2(2)+2(3)+2(4)...

another way to write this would be

dƩk[itex]^{2}[/itex]/dn=2(1+2+3+4...)

Or...

dƩk[itex]^{2}[/itex]/dn=2(ƩK)

So since Ʃk[itex]^{2}[/itex] is (2n[itex]^{3}[/itex]+3n[itex]^{2}[/itex]+n)/6 it stands to reason that the derivative of this equation would be equal to 2(ƩK), but it isn't.

Ʃk[itex]^{2}[/itex]=(2n[itex]^{3}[/itex]+3n[itex]^{2}[/itex]+n)/6=

Ʃk[itex]^{2}[/itex]=(1/3)n[itex]^{3}[/itex]+(1/2)n[itex]^{2}[/itex]+(1/6)n=

dƩk[itex]^{2}[/itex]/dn=(n[itex]^{2}[/itex]+n+1/6) Which we said equals 2Ʃk

dƩk[itex]^{2}[/itex]/dn=(n[itex]^{2}[/itex]+n+1/6)=2Ʃk

dƩk[itex]^{2}[/itex]/dn=(n[itex]^{2}[/itex]+n+1/6)/2=Ʃk

We earlier said that Ʃk=(n[itex]^{2}[/itex]+n)/2 Which renders the above equation untrue.

The funny part of this is that this holds true for Ʃk[itex]^{3}[/itex] and a few others I have tried. Can anyone explain why this doesn't work?