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

(Solution)
m (Solution 1)
Line 14: Line 14:
 
  We need to evaluate  
 
  We need to evaluate  
 
<cmath>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</cmath>
 
<cmath>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</cmath>
We replace <math>k\cdotk!</math> with <math>((k+1)-1)\cdotk!</math>
+
We replace <math>k\cdot k!</math> with <math>((k+1)-1)\cdot k!</math>
 
<cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath>
 
<cmath>(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!</cmath>
 
Distribution yields
 
Distribution yields
<cmath>(2\cdot 1!-1\cdot1!+3\cdot2!-1\cdot2!+\cdots+n(n-1)!-1(n-1)!+(n+1)n!-1\cdotn!</cmath>
+
<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,
 
Simplifying,
 
<cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath>
 
<cmath>2!-1!+3!-2!+\cdots+n!-(n-1)!+(n+1)!-n!</cmath>

Revision as of 08:59, 3 December 2015

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 1

We need to evaluate 

\[1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!\] We replace $k\cdot k!$ with $((k+1)-1)\cdot k!$ \[(2-1)\cdot 1!+(3-1)\cdot 2!+(4-1)\cdot 3!+\cdots+((n)-1)(n-1)!+((n+1)-1)\cdot n!\] 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

\[(n+1)!-1!=\box((n+1)!-1)\] (Error compiling LaTeX. Unknown error_msg)
1969 Canadian MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 7 8 Followed by
Problem 7