Question 229820
<font face="Garamond" size="+2">


*[tex \LARGE \forall\,n\,>\,0,\,n\,\in\,\I]


prove


*[tex \LARGE \sum_{i=1}^n\,i^2\ =\ \frac{n(n+1)(2n+1)}{6}]


1.  Show that


*[tex \LARGE \sum_{i=1}^n\,i^2\ =\ \frac{n(n+1)(2n+1)}{6}]


is true for


*[tex \LARGE n\ =\ 1]


*[tex \ \ \ \ \ \ \ \ \ \LARGE \sum_{i=1}^1\,i^2\ =  1^2\ =\ 1]


and


*[tex \ \ \ \ \ \ \ \ \ \LARGE \frac{1(1+1)(2(1)+1)}{6}\ =\ \frac{6}{6}\ = 1]


2.  Assume


*[tex \LARGE \sum_{i=1}^n\,i^2\ =\ \frac{n(n+1)(2n+1)}{6}]


is true for


some positive integer *[tex \LARGE n], then show


the relationship is true for *[tex \LARGE n + 1], namely that:


*[tex \ \ \ \ \ \ \ \ \ \LARGE \sum_{i=1}^{n+1}\,i^2\ =\ \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}]


First note that:


*[tex \ \ \ \ \ \ \ \ \ \LARGE \sum_{i=1}^{n+1}\,i^2\ =\ \sum_{i=1}^n\, i^2 +\ (n + 1)^2]


which can be written:


*[tex \ \ \ \ \ \ \ \ \ \LARGE \sum_{i=1}^{n+1}\,i^2\ =\ \frac{n(n+1)(2n+1)}{6}\ +\ (n + 1)^2]


because we assumed the relationship to be true for some positive integer *[tex \LARGE n].


But


*[tex \ \ \ \ \ \ \ \ \ \LARGE \frac{n(n+1)(2n+1)}{6}\ +\ (n + 1)^2\ =\ \frac{2n^3+9n^2+13x+6}{6}]


(Verification of the previous step is left as an exercise for the student)


And


*[tex \ \ \ \ \ \ \ \ \ \LARGE \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}\ =\ \frac{2n^3+9n^2+13x+6}{6}]


(Verification of the previous step is also left as an exercise for the student)


Therefore, if the statement is true for some *[tex \LARGE n], it must be true for *[tex \LARGE n + 1].  Since it was proven true for *[tex \LARGE n = 1], it must be true for *[tex \LARGE n = 2], then it must be true for *[tex \LARGE n = 3]...


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>