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I need to make the proof for this

1:$$1^2 + 2^2 + 3^2 + ... + n^2=\frac{(n(n+1)(2n+1))}{6}$$

By mathematical induction I know that,

If P(n) is true for $n>3^2$ then P(k) is also true for k=N and also P(k+1) must be true.

Then,

2:$$1^2 + 2^2 + 3^2 + ... + k^2=\frac{(k(k+1)(2k+1))}{6}$$

Substitution:

3:$$1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2=\frac{((n+1)((n+1)+1)(2(n+1)+1))}{6}$$

Questions

a: Do anyone see notation errors in this post?

b: It is right so far?

Michael Rozenberg
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Haizum Skallah
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    I don't fully understand what you did with the $k$'s, but in the last part it seems like you assumed what was to be shown instead of relying on induction. – Jonathan Y. Aug 20 '13 at 20:40
  • Possible duplicate of: http://math.stackexchange.com/questions/461329/mathematical-induction-supported-by-geometric-interpretation – Amzoti Aug 20 '13 at 20:40
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    Perhaps we should go over the induction schema (we'll avoid ZF terminology): if a property (say equation of formulas) holds for some positive integer $n$, and it can be shown that whenever it holds for a positive integer $k$ it then must also hold for $k+1$, then the property holds for all positive integers greater than $n$. – Jonathan Y. Aug 20 '13 at 20:43

1 Answers1

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For the third step, notice that $$ 1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2 = \frac{(n(n+1)(2n+1))}{6}+(n+1)^2 $$

$$ = \frac{(n(n+1)(2n+1))}{6}+\frac{6(n+1)^2}{6}$$

$$ = \frac{(n+1)(n(2n+1)+6(n+1))}{6}=\dots $$

Can you finish it?

Mhenni Benghorbal
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