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rainboi

Member

Registered: 2011-05-15

Posts: 3

Formula For the Sum Of the First N Squares

Hi all,

I refer to "http://mathforum.org/library/drmath/view/56920.html"

I followed the steps on how I should proof

But im stucked at

"1.  All of the cubes cancel except for (n+1)^3"

I have no idea why this step is needed. Can anyone help me please? Thank you.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Formula For the Sum Of the First N Squares

hi rainboi

A few lines above it says

now add all these up in columns.

On the LHS you get the sum of all the cube numbers up to and including

(n+1)^3

On the right hand side you get all the cubes from 0 cubed up to n cubed.  As 0 cubed is 0 the cubes on the RHS are the same as on the Left except for the one you are asking about.

By cancelling all the other cubes you end up with some algebra that, hopefully you can handle.  The summation of the first 'n' numbers can be reduced to formula.  (Do you know it?)  The sum of squares is what you want and the sum of cubes have cancelled.

By re-arranging the terms you can construct the formula for the sum of squares.

Post back if you want the details.

Bob

Last edited by Bob (2011-05-15 22:55:13)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

rainboi

Member

Registered: 2011-05-15

Posts: 3

Re: Formula For the Sum Of the First N Squares

1^3 =  (0 + 1)^3   = 0^3     +  3 ( 0^2 )   +  3 (0)    + 1
2^3 =  (1 + 1)^3   = 1^3     +  3 ( 1^2 )   +  3 (1)    + 1
3^3 =  (2 + 1)^3   = 2^3     +  3 ( 2^2 )   +  3 (2)    + 1
4^3 =  (3 + 1)^3   = 3^3     +  3 ( 3^2 )   +  3 (3)    + 1
            etc.
n^3 =  (n-1 + 1)^3 = (n-1)^3 +  3 (n-1)^2   +  3 (n-1)  + 1

(n+1)^3 =  (n + 1)^3   = n^3     +  3  n^2      +  3  n     + 1
----------------------------------------------------------------------

Hi bob,

Im sorry i couldnt understand

As 0 cubed is 0 the cubes on the RHS are the same as on the Left except for the one you are asking about.

im not sure which one you are referring to.

They wanted me to cancel out n^3 (in bold) but since

(n + 1)^3   = n^3     +  3  n^2      +  3  n     + 1

, wouldnt cancelling n^3 makes it not balanced?

Thanks!

Last edited by rainboi (2011-05-16 00:01:43)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Formula For the Sum Of the First N Squares

hi rainboi

Go back to here:

1^3                                       =           0^3                   +          3 ( 0^2 )        +  3 (0)                       + 1
2^3                                       =           1^3                   +          3 ( 1^2 )        +  3 (1)                       + 1
3^3                                       =           2^3                   +          3 ( 2^2 )        +  3 (2)                       + 1
4^3                                       =           3^3                   +          3 ( 3^2 )        +  3 (3)                       + 1
                                            ................................................
                                            ................................................

n^3                                       =         (n-1)^3               +          3 (n-1)^2       +  3 (n-1)                    + 1
(n+1)^3                                 =           n^3                  +          3  n^2           +  3  n                         + 1

I've spaced out the terms to make the next step easier to follow. 
Hopefully my special column allignment will remain when you view it.

There are four columns of similar terms here.  Each column is added to make one 'super equation' thus:

sum of cubes up to (n+1)^3    =   sum of cubes up to n^3 + 3 x sum of squares + 3 x sum of numbers + sum of n+1  "1" s.

Now, all the cube terms that occur on the left and on the right can be cancelled, and you are left with:

(n+1)^3                                = no cubes as they've all cancelled + 3 x sum of squares + 3 x sum of numbers + n+1 lots of "1"

so now to re-arrange

3 x sum of squares                = (n+1)^3  - 3 x sum of numbers - (n+1)

3 x sum of squares                = n^3 + 3n^2 + 3n + 1 - 3n(n+1)/2 - n - 1

double to get rid of fractions

6 x sum of squares                = 2 x n^3 + 6 x n^2 + 6n + 2 - 3n(n+1) - 2n - 2
                                           = 2 x n^3 + 6 x n^2 + 6n + 2 - 3n^2 - 3n - 2n - 2
                                           = 2 x n^3 + 3 x n^2 + n
                                           = n(2 x n^2 + 3 x n + 1)
                                           = n(n+1)(2n+1)

hence sum of squares = n(n+1)(2n+1)/6

Will that do?

Bob

Last edited by Bob (2011-05-16 00:52:38)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

rainboi

Member

Registered: 2011-05-15

Posts: 3

Re: Formula For the Sum Of the First N Squares

1^3+2^3+...+n^3+(n+1)^3  = 0^3+1^3+2^3+3^3+...+n^3+...
  \______  _____/                                 \_______  ________/
              \/                                                         \/
                these terms      cancel            these terms            .

So only the (n+1)^3 is left on the left side of the equation.

Thanks for your help!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Formula For the Sum Of the First N Squares

You are welcome!

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob