Let n! = 1 x 2 x 3 x 4 x ... x n for integer n >_1. What could be the value of 1! +(2 x 2!) + (3 x 3!) + (4 x 4!) + ... (10 x 10!)?

39,916,799

http://www.wolframalpha.com/input/?i=sum+n*n!+for+n%3D1+to+10

To find the value of 1! + (2 x 2!) + (3 x 3!) + (4 x 4!) + ... + (10 x 10!), we can first calculate each term separately and then add them together.

Let's start with the first term: 1!

1! is equal to 1.

Now, let's move on to the second term: (2 x 2!)

2! is equal to 2 x 1 = 2. Therefore, (2 x 2!) = (2 x 2) = 4.

Continuing this pattern, we can find the values of the next terms:

(3 x 3!) = (3 x 6) = 18
(4 x 4!) = (4 x 24) = 96

You can notice a pattern here. If we look closely, we can see that each term is the product of the corresponding number and the factorial of that number. For example, the nth term is given by n x n!, where n is the corresponding number.

Using this pattern, let's calculate the remaining terms:

(5 x 5!) = (5 x 120) = 600
(6 x 6!) = (6 x 720) = 4320
(7 x 7!) = (7 x 5040) = 35280
(8 x 8!) = (8 x 40320) = 322560
(9 x 9!) = (9 x 362880) = 3265920
(10 x 10!) = (10 x 3628800) = 36288000

Now, let's sum up all these terms:

1! + (2 x 2!) + (3 x 3!) + (4 x 4!) + (5 x 5!) + (6 x 6!) + (7 x 7!) + (8 x 8!) + (9 x 9!) + (10 x 10!)
= 1 + 4 + 18 + 96 + 600 + 4320 + 35280 + 322560 + 3265920 + 36288000

Adding all the terms gives us the final result:

1 + 4 + 18 + 96 + 600 + 4320 + 35280 + 322560 + 3265920 + 36288000
= 39916821

Therefore, the value of 1! + (2 x 2!) + (3 x 3!) + (4 x 4!) + ... + (10 x 10!) is 39916821.