Difference between revisions of "1969 Canadian MO Problems/Problem 6"

Revision as of 19:32, 27 November 2023

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 1

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 $(n+1)!-1.$

Solution 2

We need to evaluate: $1\cdot 1!+2\cdot 2!+\cdots+(n-1)(n-1)!+n\cdot n!$ We replace $k\cdot k!$ with $((k+1)-1)\cdot k!$ $(2-1)\cdot 1!+(3-1)\cdot 2!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!$ Distribution yields $(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdot n!$ Simplifying, $2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!$ Which telescopes to $(n+1)!-1!=(n+1)!-1$ So $(n+1)!-1$ is the solution.

Solution 3

$S=\sum_{k=1}^{n}k\cdot k!=\sum_{k=1}^{n}(k\cdot k!+k!-k!)=\sum_{k=1}^{n}\left( (k+1)\cdot k!-k! \right)=\sum_{k=1}^{n}(k+1)\cdot k!-\sum_{k=1}^{n}k!\\ =\sum_{k=2}^{n+1}k!-\sum_{k=1}^{n}k!=(n+1)!+\sum_{k=2}^{n}k!-\sum_{k=2}^{n}k!-1!=\boxed{(n+1)!-1}$

~Tomas Diaz. orders@tomasdiaz.com

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

@@ Line 24: / Line 24: @@
 <cmath>(n+1)!-1!=(n+1)!-1</cmath>
 So <math>(n+1)!-1</math> is the solution.
+== Solution  3==
+<math>S=\sum_{k=1}^{n}k\cdot k!=\sum_{k=1}^{n}(k\cdot k!+k!-k!)=\sum_{k=1}^{n}\left( (k+1)\cdot k!-k! \right)=\sum_{k=1}^{n}(k+1)\cdot k!-\sum_{k=1}^{n}k!\\
+=\sum_{k=2}^{n+1}k!-\sum_{k=1}^{n}k!=(n+1)!+\sum_{k=2}^{n}k!-\sum_{k=2}^{n}k!-1!=\boxed{(n+1)!-1}</math>
+~Tomas Diaz. orders@tomasdiaz.com
 {{Old CanadaMO box|num-b=5|num-a=7|year=1969}}

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