Calculating the Sum of a Factorial Series

The question asks for the value of a specific mathematical series involving factorials. Let's break down the series and identify the pattern.

The series is given as:

2(2 × 1) + 3(3 × 2× 1 ) + 4(4 × 3×  2×  1) + 5(5 ×  4×  3  ×  2×  1) + .................. + 9(9 ×  8×  7×  6×  5×  4×  3×  2×  1) + 2

We can rewrite the terms using factorial notation. Recall that $n! = n \times (n-1) \times \dots \times 2 \times 1$.

• The first term is $2 \times (2 \times 1) = 2 \times 2!$
• The second term is $3 \times (3 \times 2 \times 1) = 3 \times 3!$
• The third term is $4 \times (4 \times 3 \times 2 \times 1) = 4 \times 4!$
• ...
• The last term in the sum part is $9 \times (9 \times 8 \times \dots \times 1) = 9 \times 9!$

So, the series can be written as the sum of terms of the form $n \times n!$ from $n=2$ to $n=9$, plus the final constant term 2.

The sum part is $\sum_{n=2}^{9} n \times n!$.

Simplifying the General Term $n \times n!$

Let's look at the general term $n \times n!$. We can try to express this in terms of factorials of $n$ and $n+1$.

Consider the factorial $(n+1)!$. By definition, $(n+1)! = (n+1) \times n!$.

Now, let's look at the difference between $(n+1)!$ and $n!$:

$(n+1)! - n! = (n+1)n! - n!$

We can factor out $n!$ from the terms on the right side:

$(n+1)! - n! = n!( (n+1) - 1 )$

$(n+1)! - n! = n!(n)$

So, we find that $n \times n! = (n+1)! - n!$. This is a very useful identity for summing series involving $n \times n!$.

Applying the Identity to the Series Sum

Now we substitute this identity into each term of our series sum $\sum_{n=2}^{9} n \times n!$:

• For $n=2$: $2 \times 2! = (2+1)! - 2! = 3! - 2!$
• For $n=3$: $3 \times 3! = (3+1)! - 3! = 4! - 3!$
• For $n=4$: $4 \times 4! = (4+1)! - 4! = 5! - 4!$
• For $n=5$: $5 \times 5! = (5+1)! - 5! = 6! - 5!$
• For $n=6$: $6 \times 6! = (6+1)! - 6! = 7! - 6!$
• For $n=7$: $7 \times 7! = (7+1)! - 7! = 8! - 7!$
• For $n=8$: $8 \times 8! = (8+1)! - 8! = 9! - 8!$
• For $n=9$: $9 \times 9! = (9+1)! - 9! = 10! - 9!$

The sum of these terms is:

$(3! - 2!) + (4! - 3!) + (5! - 4!) + (6! - 5!) + (7! - 6!) + (8! - 7!) + (9! - 8!) + (10! - 9!)$

This is a telescoping series. Notice that most terms cancel out:

$3!$ cancels with $-3!$

$4!$ cancels with $-4!$

... and so on, until $9!$ cancels with $-9!$.

Only the first part of the first term and the second part of the last term remain:

Sum of the series part $= -2! + 10!$

Sum of the series part $= 10! - 2!$

Including the Final Constant Term

The original question includes an additional '+ 2' at the end.

So, the total value is $(10! - 2!) + 2$.

We know that $2! = 2 \times 1 = 2$.

Total value $= (10! - 2) + 2$

Total value $= 10! - 2 + 2$

Total value $= 10!$

Conclusion

The value of the given series is $10!$. Let's check the options provided:

• Option 1: $11!$
• Option 2: $10!$
• Option 3: $10 + 10!$
• Option 4: $11 + 10!$

Our calculated value $10!$ matches Option 2.

The final answer is $10!$.

Revision Table: Key Concepts for Factorial Series

Additional Information: Exploring Factorial Properties and Series Sums

Understanding factorials and their properties is crucial for solving problems like this. The identity $n \times n! = (n+1)! - n!$ is a powerful tool for summing series involving terms of the form $n \times n!$. This identity allows us to transform the series into a telescoping sum, where cancellation simplifies the calculation greatly.

Let's look at a slightly more general sum using this identity:

$\sum_{n=1}^{N} n \times n! = \sum_{n=1}^{N} ( (n+1)! - n! )$

Writing out the terms:

$(2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((N+1)! - N!)$

This sum telescopes to $(N+1)! - 1!$. Since $1! = 1$, the sum is $(N+1)! - 1$.

In our specific problem, the sum started from $n=2$ and went up to $n=9$. The sum of the series part was $\sum_{n=2}^{9} n \times n! = \sum_{n=2}^{9} ( (n+1)! - n! ) = (3! - 2!) + (4! - 3!) + \dots + (10! - 9!)$. This also telescopes, leaving the last positive term and the first negative term: $10! - 2!$.

The final '+ 2' in the original question then precisely cancels out the $-2!$ term, simplifying the total sum to just $10!$. This shows how a seemingly complex series can be solved elegantly by recognizing patterns and using mathematical identities.