Find the sum
`3/(1!+2!+3!)+4/(2!+3!+4!)+...+1000/(998!+999!+1000!)`

Explanation

To find the sum of the series

\[ S = \frac{3}{1! + 2! + 3!} + \frac{4}{2! + 3! + 4!} + \ldots + \frac{1000}{998! + 999! + 1000!}, \]

we can express the \(n\)-th term of the series.

### Step 1: Identify the \(n\)-th term

The \(n\)-th term can be expressed as:

\[ T_n = \frac{n + 2}{n! + (n + 1)! + (n + 2)!}. \]

### Step 2: Simplify the denominator

We can factor out \(n!\) from the denominator:

\[ T_n = \frac{n + 2}{n! \left(1 + (n + 1) + (n + 2)(n + 1)\right)}. \]

Calculating the terms inside the parentheses:

\[ 1 + (n + 1) + (n + 2)(n + 1) = 1 + (n + 1) + (n^2 + 3n + 2) = n^2 + 4n + 4. \]

Thus, we have:

\[ T_n = \frac{n + 2}{n! (n^2 + 4n + 4)} = \frac{n + 2}{n! (n + 2)^2}. \]

### Step 3: Cancel terms

This simplifies to:

\[ T_n = \frac{1}{n! (n + 2)}. \]

### Step 4: Rewrite the term

We can rewrite \(T_n\) as:

\[ T_n = \frac{1}{(n + 1)!} - \frac{1}{(n + 2)!}. \]

### Step 5: Write the sum

Now, we can write the sum \(S\) as:

\[ S = \sum_{n=1}^{998} \left( \frac{1}{(n + 1)!} - \frac{1}{(n + 2)!} \right). \]

### Step 6: Recognize the telescoping series

This is a telescoping series. When we expand it, we get:

\[ S = \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{3!} - \frac{1}{4!} \right) + \ldots + \left( \frac{1}{999!} - \frac{1}{1000!} \right). \]

Most terms cancel out, leaving us with:

\[ S = \frac{1}{2!} - \frac{1}{1000!}. \]

### Step 7: Final result

Calculating the final result:

\[ S = \frac{1}{2} - \frac{1}{1000!}. \]

### Conclusion

Thus, the sum of the series is:

\[ S = \frac{1}{2} - \frac{1}{1000!}. \]