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

Latest revision as of 19:33, 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

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Find the sum of <math>\displaystyle 1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math>\displaystyle  n!=n(n-1)(n-2)\cdots2\cdot1</math>.
+Find the sum of <math>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math> n!=n(n-1)(n-2)\cdots2\cdot1</math>.
-== Solution ==
+== Solution  1==
-Note that for any positive integer <math>\displaystyle  n,</math> <math>\displaystyle  n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.</math>
+Note that for any [[positive integer]] <math> n,</math> <math> n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.</math>
 Hence, pairing terms in the series will telescope most of the terms.
-If <math>\displaystyle  n</math> is odd, <math>\displaystyle  (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math>
+If <math> n</math> is [[odd integer | odd]], <math> (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math>
-If <math>\displaystyle  n</math> is even, <math>\displaystyle  (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.</math>
+If <math> n</math> is [[even integer | even]], <math> (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.</math>
-In both cases, the expression telescopes into <math>\displaystyle  (n+1)!-1.</math>
+In both cases, the expression telescopes into <math> (n+1)!-1.</math>
-----
+== Solution  2==
-* [[1969 Canadian MO Problems/Problem 5|Previous Problem]]
-* [[1969 Canadian MO Problems/Problem 7|Next Problem]]
+We need to evaluate:
-* [[1969 Canadian MO Problems|Back to Exam]]
+<cmath>1\cdot 1!+2\cdot 2!+\cdots+(n-1)(n-1)!+n\cdot n!</cmath>
+We replace <math>k\cdot k!</math> with <math>((k+1)-1)\cdot k!</math>
+<cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath>
+Distribution yields
+<cmath>(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdot n!</cmath>
+Simplifying,
+<cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath>
+Which telescopes to
+<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
+{{alternate solutions}}
+{{Old CanadaMO box|num-b=5|num-a=7|year=1969}}

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