109k views
4 votes
Prove that 1 · 1! 2 · 2! ··· n · n! = (n 1)! - 1 whenever n is a positive integer.

User Fried Rice
by
8.7k points

1 Answer

3 votes

Final answer:

The expression 1 · 1! 2 · 2! … n · n! equals (n+1)! - 1 because each term in the original series represents the factorial of the next number, leading to successive cancellations that result in (n+1)! minus the uncanceled 1.

Step-by-step explanation:

To prove that 1 · 1! 2 · 2! … n · n! equals (n+1)! - 1, we must first recognize that n! represents n factorial, which is the product of all positive integers from 1 to n. A key observation here is that n · n! is the same as (n+1)! because of the factorial property. Now consider the expression:

  • 1 · 1!
  • 2 · 2!
  • n · n!

Each term in this sequence is essentially the factorial of the next number. For instance, 2 · 2! is equivalent to 3!, and in general, a · a! = (a+1)!. This means we can rewrite this expression as:

  • 2!
  • 3!
  • (n+1)!

Now, we add all these together, but notice that there is a pattern where each factorial from 2! to n! is cancelled out successively by the term before it in the expression. This successive cancellation leaves us with just (n+1)! minus the initial 1 that was not cancelled. So we have:

(n+1)! - 1

Which is exactly what we set out to prove.

User Wilbur
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories