Question

Any hints on how to do this?

Prove that \[1^2+2^2+...+n^2 = \frac{ 1 }{ 6 }n(n+1)(2n+1)\] for all \[n \epsilon N\]

Answer 1

http://jeremykun.files.wordpress.com/2011/06/triangle-proof.png

Answer 2

proof by induction
direct proof
visual proof

Answer 3

Ok, so, Im doing this by induction. Step 1, Ive proven its true for P(1). Step 2, Assume true for P(n). Step 3 is where I am stuck. Do I simply replace all the "n"s with "n+1"

Answer 4

so proof by induction

Answer 5

I cant seem to make the right side look like the left, which is what I thought you do with induction..

Answer 6

u must prove P(n+1) is true\[1+2^2+3^2+...+(n+1)^2=\frac{(n+1)(n+2)(2n+3)}{6}\]using that P(n) is true\[1+2^2+3^2+...+n^2=\frac{(n)(n+1)(2n+1)}{6}\]

Answer 7

hint\[1+2^2+3^2+...+(n+1)^2=1+2^2+3^2+...+n^2+(n+1)^2\]\[=\frac{n(n+1)(2n+1)}{6}+(n+1)^2\]so u just need to prove\[\frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{(n+1)(n+2)(2n+3)}{6}\]

Answer 8

Sorry to keep beating a dead horse here, but I think something is wrong. Expanding both sides gives me \[\frac{ k^3 }{ 3 }+k^2+\frac{ k }{ 2 }+1 = \frac{ k^3 }{ 3 }+\frac{ 3k^2 }{ 2 }+\frac{ 13k }{ 6 }+1\]

Answer 9

which doesn't appear true.