Sum of Consecutive SquaresDate: 05/11/2001 at 19:59:27 From: Robert DiPietro Subject: Sum notation Hi, I have searched over and over for how this equation was derived: n Sum [i^2] = ((n)(n+1)(2n+1))/6 i=1 Why? Thank you. Date: 05/11/2001 at 20:12:35 From: Doctor Schwa Subject: Re: Sum notation Hi Robert, My favorite proof goes like this. Start with the sum of i^3 instead, for reasons that will become clear in a bit. 1^3 + 2^3 + 3^3 + ... + n^3 Now let's compare that with: 0^3 + 1^3 + 2^3 + ... + (n-1)^3 On the one hand, if you subtract the equal numbers, you have n^3 and all the rest of the numbers drop out. On the other hand, if you subtract vertically, you have (1^3 - 0^3) + (2^3 - 1^3) + ... + (n^3 - (n-1)^3), or in other words, sum from i = 1 to n of (i^3 - (i-1)^3). So that sum equals n^3. Multiplying that out, it's the same as summing from i = 1 to n of (3i^2 - 3i + 1), which still equals n^3. That is: 3 * Sum [i^2] - 3 * Sum [i] + Sum [1] = n^3 3 * Sum [i^2] - 3 * (n)(n+1)/2 + n = n^3 3 * Sum [i^2] = n^3 + 3n(n+1)/2 - n Sum [i^2] = (1/3) [n^3 + 3n(n+1)/2 - n] Simplifying the expression on the right side gives you just what you were looking for. This trick works in general: if you know how to sum 1, i and i^2, you can use it to figure out how to sum i^3, and so on. There are lots of other methods, though, and you might find some of them by searching our Ask Dr. Math archives at: Search Dr. Math http://mathforum.org/mathgrepform.html I found one explanation that gives a slightly different version of the approach: Formula For the Sum Of the First N Squares http://mathforum.org/dr.math/problems/sandin2.20.98.html Two different approaches, finite differences and undetermined coefficients you might call them, are at: Finding sum formula using sequences of differences http://mathforum.org/dr.math/problems/kyungsoo6.28.98.html Enjoy, - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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