Question

prove by induction that 1^2 + 2^2 + 3^2 ... + n^2 = n(n+1)(2n+1)/6

Answer 1

Base case: n=1
1^2 =? 1*(1+1)*(2*1+1)/6
the left hand side is 1
the right hand side is 1*2*3/6 = 6/6 = 1.
so n=1 works.

Now assume the formula works for numbers up to n. Does it work for n+1?
start with
1^2 + .... n^2 + (n+1)^2
now we assumed the formula works for the numbers up to n, so rewrite this sum as
n(n+1)(2n+1)/6 + (n+1)^2
n(n+1)(2n+1)/6 + 6(n+1)(n+1)/6
[n(2n+1) + 6n+6](n+1)/6
[ 2n*n + 7n + 6](n+1)/6
(n+2)(2n+3)(n+1)/6
rename n+1 as m to make the answer more obvious
note: (2n+3)= (n+n+1+1+1)= (n+1)+(n+1)+1= m+m+1= 2m+1
(m+1)(2m+1)m/6
which is the original formula

Answer 2

so m = n + 1?

Answer 3

yes, I only did that so we can compare to the original formula
n(n+1)(2n+1)/6

Answer 4

i see, Thanks alot!

Answer 5

Turing test wrote it out nicely in the equation editor in your other post

Answer 6

oh, ok