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August 14th, 2014, 05:50 PM  #1 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  Deriving the formula for summing a sequence of squares
If we have the sequence $\displaystyle 1^{2}, 2^{2}, 3^{2}, 4^{2},\cdots n^{2}$, how do we derive $\displaystyle \sum_{k = 1}^{n}k^{2}$ for any positive integer $\displaystyle n$?

August 14th, 2014, 06:01 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
The general solution for this problem is complex and involves Bernoulli numbers. But if you know that the sum of an nth degree polynomial is an (n+1)th degree polynomial, you can do a simple curve fit through the points (0, 0), (1, 1), and (2, 5).

August 14th, 2014, 06:09 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,328 Thanks: 2451 Math Focus: Mainly analysis and algebra 
Hi Mr. Davis, I think you posted a question on sequences for which the answer was $n(n+1)$. The method of differences that I used there will serve you well here too. 
August 14th, 2014, 08:39 PM  #5 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  But the third degree polynomial will require four different points for curvefitting.

August 14th, 2014, 08:53 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I did not mean (2, 4), I meant (2, 5). But you're right, you need another point. Take (3, 14).

August 14th, 2014, 10:01 PM  #7 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  
August 15th, 2014, 03:29 AM  #8 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
The question is indeed about the sum of the (first lowest) n squares, which are $\displaystyle 0, 1^2 = 1, 1^2 + 2^2 = 5, 1^2 + 2^2 + 3^2 = 14,...$. To find them for some n I'd write $\displaystyle k^2 = 2 \cdot {k \choose 2} + {k \choose 1}$ and then $\displaystyle \sum_{k=1}^{n}k^2 = \sum_{k=1}^{n}\left(2 \cdot {k \choose 2} + {k \choose 1}\right) = 2 \cdot {n+1 \choose 3} + {n+1 \choose 2}$ 
August 15th, 2014, 03:22 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 19,168 Thanks: 1640 
$\displaystyle \text{As }6k^2 \equiv 2(k^3  (k  1)^3) + 3(k^2  (k  1)^2) + 1,$ $\displaystyle 6\sum_{k=1}^n k^2 = \sum_{k=1}^n (2(k^3  (k  1)^3) + 3(k^2  (k  1)^2) + 1) = 2n^3 + 3n^2 + n = n(n + 1)(2n + 1).$ 
August 16th, 2014, 05:02 PM  #10  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, Mr Davis 97! I'll show you a very primitive solution to this problem. Quote:
We have: $\:S_n \;=\;1^2+2^2+3^2+4^2+\cdots + n^2$ Crank out the first few partial sums. $\qquad\begin{array}{ccccccc}S_1 &=& 1^2 &=& 1 \\ S_2 &=& 1^2+2^2 &=& 5 \\\ S_3 &=& 1^2+2^2+3^2 &=& 14 \\ S_4 &=& 1^2+2^2+3^2+4^2 &=& 30 \\ S_5 &=& 1^2+2^3+3^2+4^2+5^2 &=& 55 \end{array}$ Take the differences of consecutive terms, $\quad$then take the differences of the differences, and so on. $\begin{array}{cccccccccccc} S_n & 1 && 5 && 14 && 30 && 55 && 91 \\ \hline \text{1st diff} && 4 && 9 && 16 && 25 && 36 \\ \text{2nd diff} &&& 5 && 7 && 9 && 11 \\ \text{3rd diff} &&&& 2 && 2 && 2 \end{array}$ The 3rd differences are constant. Hence, the generating function is of the 3rd degree (a cubic). The general cubic is: $\:f(n) \:=\:an^3 + bn^2 + cn + d$ Use the first four values from our list: $\quad \begin{array}{ccccc}f(1) \:=\:1\!: & a + b + c + d &=& 1 \\ f(2) \:=\:5\!: & 8a + 4b + 2c + d &=& 5 \\ f(3) \:=\:14\!: & 27a + 9b + 3c + d &=& 14 \\ f(4)\:=\:30\!: & 64a + 16b + 4c + d &=& 30 \end{array}$ Solve the system: $\:a = \frac{1}{3},\;b = \frac{1}{2},\;c = \frac{1}{6},\;d = 0$ Hence: $\:f(n) \;=\;\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \;=\;\frac{1}{6}n(2n^2 + 3n + 1) $ Therefore: $\displaystyle\:\sum^n_{k=1} k^2 \;=\;\frac{n(n+1)(2n+1)}{6}$  

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deriving, formula, sequence, squares, summing 
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