a maths problem (1 Viewer)

[Real Madrid]

Theres arithmetic series and geometric series...


well..................now theres quadratic series.

Out of my own free will I wanted to investigate a series question which is

4(Sigma)n=2=n^2

Which basically is 2^2+3^2+4^2=29

...now, is there any way to plug in a formula to find the "common difference".
I've experimented with different Series sums and I found that the common difference was (2n+5)/3 for a series of n^2, when treated as an arithmetic series Sn= n/2(2a+(n-1)d) >>> n/2(2a+{[(n-1)(2n+5)]/3}) for n^2.

Don't believe me?

Find the sum of the first 15 terms: n^2

Analogue (high school method):

(1)^2+(2)^2+(3)^2+(4)^2+(5)^2+(6)^2+(7)^2+(8)^2+(9 )^2+(10)^2+(11)^2+(12)^2+(13)^2+(14)^2+(15)^2=1240

Applying Sn formula with differences:

Sn= n/2(2a+(n-1)d)
Sn=(15)/2(2(1)+((15)-1)(2n+5)/3)
Sn=7.5(2+14(2(15)+5)/3)
Sn=7.5(2+14(35/3))
Sn=7.5(2+(490/3))
Sn=7.5(496/3)
Sn=1240

Heres an example of the series under analysis:
1,2,3, 4,, 5,, 6,, 7,, 8,, 9,, 10
1,4,9,16,25,36,49,64,81,100
,,3,5,7,,9,,11,,13,15,17,19
,,,2,,2,2,,2,,,,2,,,,2,,2,,,2

As stated before, is it possible to obtain a common difference with algebraic method when a quadratic n^2 is treated as an arithmetic progression?

 

[lolokay]

sum of n square terms starting at 1, is n(n+1)(2n+1)/6

 

[Trebla]

First of all (2n + 5)/3 cannot be a "common difference" because it depends on n. If there was a common difference, it should be a constant regardless of n.
The best way to derive the formula with your current knowledge is to use collapsing sums: