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 Skip Navigation LinksMath Help > Geometry > Points and Lines > Determine A Polynomial > Finding Coefficients of the Sum from k=1 to n of k^2

Following are excerpts from a bulletin board, called the mathboard on the algebra-online.com website.

In this thread, a student is asking why it's true that the sum of squares from 1 to n^2 is equal to n(n+1)(2n+1)/6.  The first response was to show it inductively, but this left the student unsatisfied...

If I get your question, you're asking how a person would stumble over the formula in the first place; not how to prove it's correct once you know it.  In answer to that question, I would like to introduce you to a general method of figuring out what polynomial fits a given series.  This works if there is a polynomial that fits the series, but it doesn't require you to know the coefficients of that polynomial in advance.

First, make a table like the one below.  The third column, sum of n², is the one that we're going to fit to a polynomial.  The next column is the differences between successive entries of the third column.  The column after that are the successive differences of the fourth column, etc.  Eventually, if a polynomial fits the series, you'll get to a column of constant differences.  In this case, the 3rd differences are constant, so a 3rd degree polynomial fits this series.

nsum of n²1st diff2nd diff3rd diff
000   
1111  
24543 
3914952
416301672
525552592
6369136112
74914049132
86420464152
98128581172
10100385100192

Divide the constant 3rd differences by 3! (3 factorial, which is 6) to get the 3rd degree coefficient for the polynomial.  Now make a new table, subtracting this term (n³/3) from the series we're trying to fit, and find the differences all over again:

n(sum of n²)-(n³/3)1st diff2nd diff3rd diff
000   
112/32/3  
242 1/31 2/31 
3952 2/310
4168 2/33 2/310
52513 1/34 2/310
636195 2/310
74925 2/36 2/310
86433 1/37 2/310
981428 2/310
1010051 2/39 2/310

Now the 2nd differences are constant.  Divide this number by 2! to get the 2nd degree coefficient.  Then subtract this term from the series we're trying to fit, and find the differences all over again.

n(sum of n²)-(n³/3+n²/2)1st diff2nd diff3rd diff
000   
111/61/6  
241/31/60 
391/21/600
4162/31/600
5255/61/600
63611/600
7491 1/61/600
8641 1/31/600
9811 1/21/600
101001 2/31/600

Now the 1st differences are constant.  Do the same thing to get the 1st degree coefficient, subtract it, and do the differences again:

n(sum of n²)-(n³/3+n²/2+n/6)1st diff2nd diff3rd diff
000   
1100  
24000 
390000
4160000
5250000
6360000
7490000
8640000
9810000
101000000

Now you can see that the sum of  n² is equal to (n³/3+n²/2+n/6).  You can factor this to get n(n+1)(2n+1)/6.

The great thing about this method is that it will help you find the sum of n³, or any higher power of n for that matter, as well as any other series that can be exactly fit by a polynomial of finite order.

Related pages in this website

Procedures

Method of Successive Differences to find the coefficients of a polynomial f(k), given a few values of f(k) for successive integers, k.

Find the equation of an order n polynomial that passes through n points

Academic Decathlon State Competition Question 06 -- How many squares are on a checkerboard? 

 

The webmaster and author of this Math Help site is Graeme McRae.