Let's analyze the given problem. We are given a series where the \(n\)-th term is defined as \(a_n = n \cdot (n!)\). We need to find the sum of the first 10 terms of this series, which is \(a_1 + a_2 + a_3 + \dots + a_{10}\).

Understanding the Series Term a_n = n(n!)

The \(n\)-th term of the series is given by the formula \(a_n = n \cdot n!\). Let's write out the first few terms to see if we can find a pattern:

• For \(n=1\), \(a_1 = 1 \cdot 1! = 1 \cdot 1 = 1\)
• For \(n=2\), \(a_2 = 2 \cdot 2! = 2 \cdot (2 \times 1) = 2 \cdot 2 = 4\)
• For \(n=3\), \(a_3 = 3 \cdot 3! = 3 \cdot (3 \times 2 \times 1) = 3 \cdot 6 = 18\)
• For \(n=4\), \(a_4 = 4 \cdot 4! = 4 \cdot (4 \times 3 \times 2 \times 1) = 4 \cdot 24 = 96\)

We need to calculate the sum \(S = a_1 + a_2 + \dots + a_{10}\).

Finding a Simpler Expression for a_n

Let's try to rewrite the term \(a_n = n \cdot n!\) in a different form. Consider the difference between two consecutive factorial terms multiplied by related numbers. A common technique for sums involving factorials is to relate \(n \cdot n!\) to expressions involving \((n+1)!\) or \(n!\).

Let's look at \((n+1)! - n!\):

\[ (n+1)! - n! = (n+1) \cdot n! - 1 \cdot n! \]

We can factor out \(n!\) from both terms:

\[ (n+1)! - n! = n! \cdot ((n+1) - 1) \]

\[ (n+1)! - n! = n! \cdot (n) \]

So, we have discovered a very useful identity: \(n \cdot n! = (n+1)! - n!\).

This identity shows that each term \(a_n\) can be expressed as the difference of two consecutive terms involving factorials: \(a_n = (n+1)! - n!\).

Calculating the Sum using the Telescoping Series Method

Now we can write the sum of the series using this new form of \(a_n\):

\[ S = \sum_{n=1}^{10} a_n = \sum_{n=1}^{10} ((n+1)! - n!) \]

Let's write out the terms of the sum:

• \(a_1 = (1+1)! - 1! = 2! - 1!\)
• \(a_2 = (2+1)! - 2! = 3! - 2!\)
• \(a_3 = (3+1)! - 3! = 4! - 3!\)
• \(a_4 = (4+1)! - 4! = 5! - 4!\)
• ...
• \(a_{10} = (10+1)! - 10! = 11! - 10!\)

Now, let's add these terms together:

\[ S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (10! - 9!) + (11! - 10!) \]

Notice that many terms cancel out. The \(-2!\) from the first term cancels with the \(+2!\) from the second term. The \(-3!\) from the second term cancels with the \(+3!\) from the third term, and so on. This is called a telescoping sum.

\[ S = \cancel{(2!)} - 1! + \cancel{(3!)} - \cancel{(2!)} + \cancel{(4!)} - \cancel{(3!)} + \dots + \cancel{(10!)} - \cancel{(9!)} + 11! - \cancel{(10!)} \]

After cancellation, only the first part of the first term and the second part of the last term remain:

\[ S = -1! + 11! \]

Since \(1! = 1\), the sum is:

\[ S = 11! - 1 \]

Final Answer

The sum \(a_1 + a_2 + a_3 + \dots + a_{10}\) is equal to \(11! - 1\).

Revision Table: Key Concepts

Additional Information: Extending the Series

The pattern \(n \cdot n! = (n+1)! - n!\) is very useful. We can use it to find the sum of any number of terms in this series.

For example, the sum up to \(N\) terms would be:

\[ \sum_{n=1}^{N} n \cdot n! = \sum_{n=1}^{N} ((n+1)! - n!) \]

This sum would telescope to \((N+1)! - 1!\), which simplifies to \((N+1)! - 1\).

In our specific problem, \(N = 10\), so the sum is \((10+1)! - 1 = 11! - 1\).