Verify that $\dfrac{3}{(r+1)!} - \dfrac{2}{r!} - \dfrac{1}{(r-1)!} = -\dfrac{r^2 + 3r + 1}{(r+1)!}$. Hence find $\displaystyle\sum_{r=1}^{n} \dfrac{-r^2 - 3r + 1}{(r+1)!}.$ Use your answer to part (b) to find $\displaystyle\sum_{r=3}^{n} \dfrac{-r^2 - r + 3}{r!}.$

1. Replace the summand using the result of part (a)

From part (a), the summand $\dfrac{-r^2-3r+1}{(r+1)!}$ can be written as
$\frac{-r^2-3r+1}{(r+1)!}=\frac{3}{(r+1)!}-\frac{2}{r!}-\frac{1}{(r-1)!}.$
So
$\sum_{r=1}^{n}\frac{-r^2-3r+1}{(r+1)!}=\sum_{r=1}^{n}\left(\frac{3}{(r+1)!}-\frac{2}{r!}-\frac{1}{(r-1)!}\right).$

2. Express the summand as g(r) − g(r−1)

Let $g(r)=\dfrac{1}{r!}+\dfrac{3}{(r+1)!}$. Then
$g(r)-g(r-1)=\left(\frac{1}{r!}+\frac{3}{(r+1)!}\right)-\left(\frac{1}{(r-1)!}+\frac{3}{r!}\right)=\frac{3}{(r+1)!}-\frac{2}{r!}-\frac{1}{(r-1)!},$
which is exactly the summand. So the sum telescopes:
$\sum_{r=1}^{n}\big(g(r)-g(r-1)\big)=g(n)-g(0).$

3. Evaluate the telescoped expression

$g(n)=\frac{1}{n!}+\frac{3}{(n+1)!},\qquad g(0)=\frac{1}{0!}+\frac{3}{1!}=1+3=4.$
Therefore
$\sum_{r=1}^{n}\frac{-r^2-3r+1}{(r+1)!}=\frac{1}{n!}+\frac{3}{(n+1)!}-4.$

1. Perform an index substitution to match the summand with the expression in part (b).

Let $k = r - 1$, so $r = k + 1$. The expression inside the sum can be rewritten by substituting $r$:
$-r^2 - r + 3 = -(k+1)^2 - (k+1) + 3 = -k^2 - 2k - 1 - k - 1 + 3 = -k^2 - 3k + 1$
Adjusting the summation limits, when $r=3$, $k=2$, and when $r=n$, $k=n-1$. The sum becomes:
$\sum_{r=3}^{n} \dfrac{-r^2 - r + 3}{r!} = \sum_{k=2}^{n-1} \dfrac{-k^2 - 3k + 1}{(k+1)!}$

2. Express the new sum using the summation starting from $k=1$ and adjust for the missing term.

This sum is identical to the one in part (b), except it starts at $k=2$ and ends at $n-1$. We can express it by taking the full sum from $k=1$ and subtracting the first term ($k=1$):
$\sum_{k=2}^{n-1} \dfrac{-k^2 - 3k + 1}{(k+1)!} = \sum_{k=1}^{n-1} \dfrac{-k^2 - 3k + 1}{(k+1)!} - \left[ \text{term for } k=1 \right]$
The $k=1$ term is:
$\dfrac{-(1)^2 - 3(1) + 1}{(1+1)!} = \dfrac{-3}{2}$
So the expression becomes:
$\sum_{k=1}^{n-1} \dfrac{-k^2 - 3k + 1}{(k+1)!} + \dfrac{3}{2}$

3. Apply the evaluated formula from part (b) to find the final expression.

From part (b), we know $\displaystyle\sum_{r=1}^{n} \dfrac{-r^2 - 3r + 1}{(r+1)!} = \dfrac{3}{(n+1)!} + \dfrac{1}{n!} - 4$, which cleanly simplifies to:
$\dfrac{3 + (n+1)}{(n+1)!} - 4 = \dfrac{n+4}{(n+1)!} - 4$
Substitute $n-1$ into this formula for the sum up to $n-1$:
$\sum_{k=1}^{n-1} \dfrac{-k^2 - 3k + 1}{(k+1)!} = \dfrac{(n-1)+4}{((n-1)+1)!} - 4 = \dfrac{n+3}{n!} - 4$
Now add the $\dfrac{3}{2}$ from Step 2:
$\left( \dfrac{n+3}{n!} - 4 \right) + \dfrac{3}{2} = \dfrac{n+3}{n!} - \dfrac{5}{2}$