steps to inductive
1. show it's true for n=1
2. replace n with k
3. replace k with k+1 and show it is true statement
4. state that it's true
1. show it's true for n=1
1(1+1)=[1(1+1)(1+2)]/3
1(2)=[(2)(3)]/3
2=6/3
2=2
true
2. assume it's true for n=k
1*2+2*3+3*4+...+k(k+1)=[k(k+1)(k+2)]/3
3. show it's true for replaceing k with k+1
1*2+2*3+3*4+...+k(k+1)+(k+1)((k+1)+1)=[(k+1)((k+1)+1)((k+1)+2)]/3
remember from before that 1*2+2*3+3*4+...+k(k+1)=[k(k+1)(k+2)]/3
[k(k+1)(k+2)]/3+(k+1)(k+2)=[(k+1)(k+2)(k+3)]/3
use algebra to show this is true (get it to k=k or something like that)
here's the fun part
deal with left side to get to like right side
[k(k+1)(k+2)]/3+(k+1)(k+2)
make over 3, so muliptlu 2nd thing by 3/3
[k(k+1)(k+2)]/3+((k+1)(k+2)3)/3
[k(k+1)(k+2)+3(k+1)(k+3)]/3
we can undistribute the (k+1)(k+2) part
[(k+1)(k+2)(k+3)]/3
now remember the right side?
[(k+1)(k+2)(k+3)]/3=[(k+1)(k+2)(k+3)]/3
tada
true
4. state it
therefor 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3 is true for all real numbers
