Summation Induction when lower limit is not 1

Question

The question is use induction to prove that

$$\sum_{r=2}^n (r^2+r+1)r! = (n+1)^2n!-4$$

I don't understand how to even get the P1 statement since when I substitute r = 2 into the LHS and n = 1 into the RHS, I get LHS = 14 and RHS = 0.

Are my substitutions wrong?

Answer 1

HINT:

Without induction,

$$(r^2+r+1)\cdot r!=\{(r+2)(r+1)-2(r+1)-1\}\cdot r!$$ $$=(r+2)!-(r+1)!-\{(r+1)!-r!\}$$

Telescoping Series!!