# The sum of series

1. Feb 1, 2006

### heaven eye

a series such as :-

∑ from n = 1 ,until m for n

can be found with the law of the sum of arithmetical series:-

[m(2a+(m-1)d)]/2
where :-
m= the number of terms
a= the first term in the series
d= the basic arithmetical

in the pervious example :-

a=1 , d=1 , m=m

when we solve the last information in the law we find :-

∑ from n = 1 ,until m for n = m(m+1)/2

for example :-
1+2+3+4+..........+18+19+20 = 20(20+1)/2 = 210

my question is :-

what about the series :-

∑ from n = 1 ,until m for n^2 ???

it isn't an artithmetical or even a geometrical series then :-

what kind of series is it ?

and how could they find that :-
∑ from n = 1 ,until m for n^2 = [m(m+1)(2m+1)]/6 ???

for example :-

(1^2)+(2^2)+(3^2)+(4^2)+(5^2)= 5(5+1)(2*5+1)/6 = 55

and thank you

2. Feb 1, 2006

### HallsofIvy

It's not an arithmetic series, obviously. In an arithmetic series, by definition, the difference between two consecutive terms must always be the same. But in 1, 22= 4, 32= 9, 42= 16, ..., the differences are 4-1 =3, 9- 4= 5, 16- 9= 7,...

Notice, however, those differences are consecutive odd numbers: the "second differences" 5-3. 7- 5, ... are all 2. Obviously, for the sum of squares, the first difference is the squares themselves so the "third differences" are all 2. "Newton's divided difference formula" gives us
$$\sum_{i=0}^n i^2= \frac{n(n+1)(2n+1)}{6}$$

Similar things can be done for higher powers but the results get progressively more difficult.

3. Feb 15, 2006

### heaven eye

Well sorry I was late to repost cause of exams

well thank you mr.HallsofIvy for your answer, now it is more clear to me

my reagrds

4. Feb 19, 2006

### robert Ihnot

If you want to put time into it, you can assume that the sum of the integers has a leading square factor, the sum of the squares has a leading cube term, the sum of third powers, etc. (This follows from HallsofIvy above.)

We then attempt to find terms $$ax^3+bx^2+cx+d$$=S(x)

We know that S(0) = 0 and so d=0. S(1) = a+b+c=1, S(2)=8a+4b+2c=5, S(3) = 27a+9b+3c=14. So we can solve these equations by elimination of terms.

Last edited: Feb 19, 2006