Mathematical Induction

Question

Use mathematical induction to prove the following statement for every positive integer n >_1:                          2+6+12+...+n(n+1)=(n(n+1)(n+2))/3

Answer 1

2+6+12+...+n(n+1)=(n(n+1)(n+2))/3

show n=1 is true, assume original equation is true, show original equation for n+1 is true

1(1+1)=(1(1+1)(1+2))/3

2=(1(2)(3))/3

2=2

show 2+6+12+...+n(n+1)+(n+1)(n+2)=((n+1)(n+2)(n+3))/3 is true

replace 2+6+12+...+n(n+1) by (n(n+1)(n+2))/3 because we assumed it to be true

(n(n+1)(n+2))/3+(n+1)(n+2)=((n+1)(n+2)(n+3))/3

(n(n+1)(n+2))/3+3(n+1)(n+2)/3=((n+1)(n+2)(n+3))/3

[(n(n+1)(n+2))+3(n+1)(n+2)]/3=((n+1)(n+2)(n+3))/3

factor the left side

[(n+1)(n+2)][n+3]/3=((n+1)(n+2)(n+3))/3

((n+1)(n+2)(n+3))/3=((n+1)(n+2)(n+3))/3

both sides are equal