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Prove that 1 · 1! 2 · 2! ··· n · n! = (n 1)! - 1 whenever n is a positive integer.

User Fried Rice
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1 Answer

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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
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