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1969 Canadian MO Problems/Problem 6

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Problem

Find the sum of 1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!, where n!=n(n-1)(n-2)\cdots2\cdot1.

Solution

Note that for any positive integer n, n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!. Hence, pairing terms in the series will telescope most of the terms.

If n is odd, (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.

If n is even, (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!. In both cases, the expression telescopes into (n+1)!-1.


1969 Canadian MO (Problems)
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 Followed by
Problem 7


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