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

1. Express the terms over a common denominator of $(r+1)!$.

$\begin{aligned} \text{LHS} &= \frac{3}{(r+1)!}-\frac{2}{r!}-\frac{1}{(r-1)!} \\ &= \frac{3}{(r+1)!} - \frac{2(r+1)}{(r+1)!} - \frac{r(r+1)}{(r+1)!} \end{aligned}$

2. Expand and simplify the numerator to obtain the Right Hand Side (RHS).

$\begin{aligned} &= \frac{3 - (2r + 2) - (r^2 + r)}{(r+1)!} \\ &= \frac{3 - 2r - 2 - r^2 - r}{(r+1)!} \\ &= \frac{-r^2-3r+1}{(r+1)!} \\ &= \text{RHS} \end{aligned}$

1. Substitute the identity verified in part (a) into the summation.

Using the identity from part (a), we can rewrite the summation term:
$\sum_{r=1}^n \frac{-r^2-3r+1}{(r+1)!} = \sum_{r=1}^n \left( \frac{3}{(r+1)!} - \frac{2}{r!} - \frac{1}{(r-1)!} \right)$

2. List the first few and last few terms to observe the telescoping cancellation pattern.

Writing out the terms row by row to easily spot the cancellations:

3. Sum the uncancelled terms and simplify.

Adding vertically, terms involving $\frac{1}{k!}$ cancel out for $k \ge 2$ because $\frac{3}{k!} - \frac{2}{k!} - \frac{1}{k!} = 0$.

The uncancelled terms that remain are:

• At the top: $-\frac{2}{1!}$ and $-\frac{1}{0!}$ (from $r=1$), and $-\frac{1}{1!}$ (from $r=2$)
• At the bottom: $\frac{3}{n!}$ (from $r=n-1$), and $\frac{3}{(n+1)!} - \frac{2}{n!}$ (from $r=n$)

Summing these remaining terms:

1. Substitute the index of summation to relate the sum to the expression in part (b).

From part (b), we established that $\displaystyle\sum_{r=1}^n\frac{-r^2-3r+1}{(r+1)!} = \frac{n+4}{(n+1)!} - 4$.

For the target sum $\displaystyle\sum_{r=3}^n\frac{-r^2-r+3}{r!}$, we can match the denominator to the known result by substituting $k = r - 1$.
The summation limits change from $r=3$ to $k=2$, and from $r=n$ to $k=n-1$.
$\begin{aligned} \sum_{r=3}^n\frac{-r^2-r+3}{r!} &= \sum_{k=2}^{n-1}\frac{-(k+1)^2-(k+1)+3}{(k+1)!} \\ &= \sum_{k=2}^{n-1}\frac{-(k^2+2k+1)-k-1+3}{(k+1)!} \\ &= \sum_{k=2}^{n-1}\frac{-k^2-3k+1}{(k+1)!} \end{aligned}$

2. Rewrite the sum using the result from part (b) by adjusting limits.

This new sum is exactly the series from part (b), but evaluated up to $n-1$ and missing the first term ($k=1$). We can rewrite it as:
$\sum_{k=2}^{n-1}\frac{-k^2-3k+1}{(k+1)!} = \sum_{k=1}^{n-1}\frac{-k^2-3k+1}{(k+1)!} - \left( \frac{-(1)^2-3(1)+1}{(1+1)!} \right)$
Using the formula from part (b) and replacing $n$ with $n-1$:
$\sum_{k=1}^{n-1}\frac{-k^2-3k+1}{(k+1)!} = \frac{(n-1)+4}{((n-1)+1)!} - 4 = \frac{n+3}{n!} - 4$

3. Evaluate the missing term and simplify the final expression.

Calculate the value of the missing $k=1$ term:
$\frac{-1-3+1}{2!} = -\frac{3}{2}$
Substitute both parts back to find the final answer:
$\begin{aligned} \sum_{r=3}^n\frac{-r^2-r+3}{r!} &= \left( \frac{n+3}{n!} - 4 \right) - \left( -\frac{3}{2} \right) \\ &= \frac{n+3}{n!} - 4 + \frac{3}{2} \\ &= \frac{n+3}{n!} - \frac{5}{2} \end{aligned}$