Sums of powers

[psi20]

1 + 2 + 3 + ...+ n = n(n+1)/2

 

1 + 4 + 9 + ...+ n^2 = n(n+1)(2n+1)/6

 

1 + 8 + 27 + ...+n^3= (n^2)((n+1)^2)/4

 

Besides just looking at the numbers and figuring out through mathematical induction, how do you find these formulas.

 

For the first one, there's a way that goes something like this.

Sum = 1 + 2 + 3+...+(n-1)+ n

Sum = n +(n-1)+...+ 2 + 1

 

2 Sum = (n)(n+1)

 

Sum = (n)(n+1)/2

 

Can this method be applied to higher powers? If so, can you show a couple of examples?

 

I also saw a method using dots. Like for n(n+1)/2, you can form a n(n+1) rectangle of dots by placing together 2 trianglular array of dots with one dot in the top row, 2 in the second, ..., n in the nth.

 

How can you use dots to find the formula of the sum for higher powers?

[uncool]

Personally, I find that the easiest way to remember is that the sum along a diagonal of Pascal's triangle is equal to the number in the next diagonal down from the last number in the sum.

Pascal's triangle:

 

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 + 2 + 3 + 4 + 5 = 15 (so go down from the 5, and you have your sum)

1 + 4 + 10 = 15

...

Now, remember the formula for each part in Pascal's triangle:

Let r = the row number, and c = the number of columns over from either 1 (where 0 means you're using the 1)

Pascal(r,c) = r!/(c!(r-c)!)

So for the third column over, Pascal(r,2) = r*(r-1)/2

For the fourth, Pascal(r,3) = r*(r-1)*(r-2)/3*2

And so on.

-Uncool-

[Primarygun]

I have two powerful way to find out the sum mathematically.

However, I am rather lazy to show it now. I can show you few days later.

1. By adding up

2. By using Interpolarization. I don't know whether the name is correct.

It is in linear algebra.

[psi20]

uncool, those are interesting sums, but not the sums I'm looking for. The sums I'm looking for are like the sum of consecutive powers of numbers. Like 1^3 + 2^3 + 3^3= (3^2)((3+1)^2)/4

 

Primarygun I couldn't find any match on interpolarization. Care to explain what you're talking about?

[Primarygun]

uncool, those are interesting sums, but not the sums I'm looking for. The sums I'm looking for are like the sum of consecutive powers of numbers. Like 1^3 + 2^3 + 3^3= (3^2)((3+1)^2)/4

 

Primarygun I couldn't find any match on interpolarization. Care to explain what you're talking about?

Sure. Several years ago, I searched for the solution and I found method 1.

Several months ago,I read some university books and focused on a method and I found out it can be used to find the sum of power as well.

Lagrange interpolation formula. Search Chinese Remainder theorem to get the concept.

If we have some ways to find out the sum of the power is actually a x degree function

, then we can set x+1 different points to find out the Appearance of the function.

If you know how to find out the degree of a function, please tell me.

For example, The sum of 1 degree is actually a 2 degree function while sum of 2 degree is actually a 3 degree function.

Since the math tool is not very familiar to me, Sorry, I can't write it here.

[Dave]

I distinctly remember posting something on this quite a while back, but it's definately here. You can use something called the method of telescoping sums to calculate these series. Unfortunately, LaTeX is down atm and I'd prefer to write about it when it's back up (saves a lot of hassle). When it's back up, I'll post some more.

[uncool]

However, what I showed can be used to find the sums - as each of the Pascal's triangle parts is a polynomial.

So, for example, to find the sum of squares:

Pascal(x,2) = x^2/2-x/2

Pascal(x+1,3) = x(x+1)(x-1)/6 = x^2/2-x/2+(x-1)^2/2-(x-1)/2+...

x(x-1)(x+1)/3 = x^2+(x-1)^2+... - x - (x-1) - ...

x(x-1)(x+1)/3+x(x+1)/2 = x^2+(x-1)^2+...

x(x+1)(2x+1)/6 = x^2+(x-1)^2+...

-Uncool-

[Johnny5]

However' date=' what I showed can be used to find the sums - as each of the Pascal's triangle parts is a polynomial.

So, for example, to find the sum of squares:

Pascal(x,2) = x^2/2-x/2

Pascal(x+1,3) = x(x+1)(x-1)/6 = x^2/2-x/2+(x-1)^2/2-(x-1)/2+...

x(x-1)(x+1)/3 = x^2+(x-1)^2+... - x - (x-1) - ...

x(x-1)(x+1)/3+x(x+1)/2 = x^2+(x-1)^2+...

x(x+1)(2x+1)/6 = x^2+(x-1)^2+...

-Uncool-[/quote']

 

I am trying to understand what you did there.

 

Just to be sure I understood you,

 

You are saying that if I understand the mathematics of Pascal's triangle, then I could find a formula for the sum of the first N cubes too?

 

1+(2)^3+(3)^3+... N^3 = formula(N)

 

?

 

Quite a few years back, after reading something on the difference calculus by Hamming, I began to do exactly what the original poster asked.

 

I sat there figuring out formulas for the sum of squares, sum of first N cubes, etcetera. And I was able to use what I'd just learned to find formulas, and there was a pattern.

 

I think I went up to 11, or 12 and then stopped. You know...

 

1^11+(2)^11+(3)^11+4^(11)+... (N)^11= formula N

 

The thing about what I was doing though, was that in order to do 12, you needed 11, and so on.