Question 229820
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*[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]
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