Finding the Sum of Series: 1^2+2^2+3^2+4^2+5^2...

[shintashi]

I was reading book on numbers and it had an equation for

1^2+2^2+3^2+4^2+5^2...

but I worked out the equation several times with different integers and they didn't appear to work at all. Does anyone know the equation for finding the total sum if the base sum (such as 5 above) is known?

 

[mathwonk]

huh? the sum of the squares of the integers from 1 to n is well known (say to archimedes) to be (1/6)[n(n+1)(2n+1)].

e.g. if n=1 we get 1,

if n = 2 we get (1/6)(2+3+5] = 5 = 1^2 + 2^2.

if n = 3 we get (1`/6)[3(4)(7)] = 14 = 1^2 + 2^2+ 3^2.

if n=4, we get 30 = (1/6)[4(5)9].
...

im not sure i understood your base sum remark, but since this is a cubic formula it is determined by the lagrange interpolation method, by the above results for n=1,2,3,4.

i.e. there is only one cubic polynomial that comes out 1,5,14,30, when we plug in 1,2,3,4.

namely 1[(x-2)(x-3)(x-4)/(1-2)(1-3)(1-4)] + 5[(x-1)(x-3)(x-4)/(2-1)(2-3)(2-4)] + ... you get it...

 

[Gib Z]

As mathwonk already said, the sum of the squares of the integers from 1 to n is \frac{n(n+1)(2n+1)}{6}.

Thats all you want to know, if you want to get confused, read forward :

When n is infinity, the sum is actually zero :D Look up the Riemann zeta function if your interested.

 

[Edgardo]

Hi shintashi,

write down here what formula is given in your book, maybe
the formula in the book is wrong.

As you indicated in your title "square pyramid", you can indeed
derive the formula for the sum of the square numbers from 1 to n
by considering the problem in a geometric view.

1 + 4 + 9 can be interpreted as a pyramid made up of
square layers:
9-layer at the bottom
4-layer in the middle
1-layer at the top

Have a look at here: http://www.sciface.com/education/data/web/SummeDerQuadrate_images/plt1.png

Look at the section Sum of the first n squares of Natural Numbers S[Q] http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm .

Look at the bottom of the website http://did.mat.uni-bayreuth.de/geonet/beispiele/kugel3/scheib6.html.

As an exercise try to figure out the formula by using the the pyramid picture.
Hint: Start by asking yourself how many times the 1-layer fits into the pyramid.
If you substract the 1-layer from all the other layers, what is left?

You will need Gauss's formula:

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

 

[shintashi]

thank you much ^^

huh? the sum of the squares of the integers from 1 to n is well known (say to archimedes) to be (1/6)[n(n+1)(2n+1)].

e.g. if n=1 we get 1,

if n = 2 we get (1/6)(2+3+5] = 5 = 1^2 + 2^2.

if n = 3 we get (1`/6)[3(4)(7)] = 14 = 1^2 + 2^2+ 3^2.

if n=4, we get 30 = (1/6)[4(5)9].
...

im not sure i understood your base sum remark, but since this is a cubic formula it is determined by the lagrange interpolation method, by the above results for n=1,2,3,4.

i.e. there is only one cubic polynomial that comes out 1,5,14,30, when we plug in 1,2,3,4.

namely 1[(x-2)(x-3)(x-4)/(1-2)(1-3)(1-4)] + 5[(x-1)(x-3)(x-4)/(2-1)(2-3)(2-4)] + ... you get it...

(1/6)[7(7+1)(14+1)] =

7x8x15 = 140.

(1/6)[n(n+1)(2n+1)].

the book said:
1/6n(n+2)(2n+1)

as you can clearly see, the equation (page 127 of kingdom of infinite number) is flawed, perhaps a typo, but changing n+1 to n+2 screwed up all the results I got from it and was obviously wrong.