Prove by induction with summation and factorials

Question

Sorry I'm not sure how to format the text for using sum. Could someone help me out with that as well. Much appreciated. \begin{eqnarray*} \sum_{r=1}^{n} (r^2+1)r! =n(n+1)! \end{eqnarray*} for all $n \geq 1 $.

Answer 1

\begin{eqnarray*} \sum_{r=1}^{n+1} (r^2+1)r! &=& \sum_{r=1}^{n} (r^2+1)r! +((n+1)^2+1)(n+1)! \\ &=&n(n+1)!+((n+1)^2+1)(n+1)! \\&=&(n^2+3n+2)(n+1)!=(n+1)(n+2)! \end{eqnarray*}

Answer 2

Hint: Notice that $n(n+1)!-(n-1)n!=\left(n^2+1\right)n!$

Answer 3

$$\sum_{r=1}^{n} (r^2+1)r! =n(n+1)!$$ $$\sum_{r=1}^{1} (r^2+1)r! = (1^2 + 1)1! = 2 * 1 = 2 = 1(1 + 1)$$ So it is true for $n=1$. Now let's see for $n=p$ $$\sum_{r=1}^p (r^2+1)r! =p(p+1)!$$ $$\sum_{r=1}^{p+1} (r^2+1)r! = (p+1)(p+2)!$$ $$\sum_{r=1}^{p} (r^2+1)r! + ((p+1)^2 + 1)(p+1)! = (p+1)(p+1)! *(p+2)$$ $$\sum_{r=1}^{p} (r^2+1)r! + (p^2 + 2p + 2)(p+1)! = (p+1)(p+2)(p+1)!$$ $$\sum_{r=1}^{p} (r^2+1)r! + (p^2 + 2p + 2)(p+1)! = (p^2 + 3p + 2)(p+1)!$$ $$\sum_{r=1}^{p} (r^2+1)r! + (p^2 + 2p + 2)(p+1)! = (p + 3 + 2 * \frac{1}{p})p(p+1)!$$ $$1 + ((p^2 + 2p + 2)(p+1)!)/(p(p+1)!) = (p + 3 + 2\frac{1}{p})$$ $$\frac{p^2 + 2p + 2}{p} + 1 = p + 3 + 2 * \frac{1}{p}$$ $$p + 2 + 2 * \frac{1}{p} + 1 = p + 3 + 2 * \frac{1}{p}$$

Which is true, so the statement is true, proven by induction