For every n ∊ ℕ, 1·2 + 2·4 + 3·5 + . . . + n(n+2) = n(n+1)(2n+7)/6

For every n ∊ ℕ,  1·2 + 2·4 + 3·5 + . . . + n(n+2) = n(n+1)(2n+7)/6

Base case: Verify the statement for P(1) 

(1)(1+2) = (1)(1+1)(2(1)+7)/6

(1)(3) = (1)(2)(9)/6

3 = 18/6 → 3 = 3 ☑

Induction case: 

Suppose P(n) holds i.e  1·2 + 2·4 + 3·5 + . . . + n(n+2) = n(n+1)(2n+7)/6

Then P(n+1)

=  1·2 + 2·4 + 3·5 + . . . + (n+1)((n+1)+2) 

which simplifies to: 

= 1·2 + 2·4 + 3·5 + . . . + (n+1)(n+3) 

= (1·2 + 2·4 + 3·5 + . . . + n(n+2)) + (n+1)(n+3) = n(n+1)(2n+7)/6 + (n+1)(n+3)

P(n)  + (n+1)(n+3) = n(n+1)(2n+7)/6 + (n+1)(n+3)

the right hand side i.e.  n(n+1)(2n+7)/6 + (n+1)(n+3) can be simplified to 

= n(n+1)(2n+7)/6 + 6(n+1)(n+3)/6

factor out (n+1)/6 and get

= (n+1/6)(n(2n+7) + 6(n+3))

= (n+1/6)(2n^2 + 7n + 6n + 18)

=(n+1/6)((n+2)(n+9)

(n+1)(n+2)(n+9)/6 ☑