Sum of Series

Sum of Series
Find the sum of the series 1(1!)\!+2(2!)\!+3(3!)\!+\cdots+n(n!)
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
1(1!)\!+2(2!)\!+3(3!)\!+\cdots+n(n!)\!=[2(1!)\!-1!]+[3(2!)\!-2!]+[4(3!)\!-3!]+\cdots+[n(n-1)!\!-(n-1)!]+[(n+1)(n!)\!-n!]
=2!\!-1!\!+3!\!-2!\!+4!\!-3!\!+\cdots+n!\!-(n-1)!\!+(n+1)!\!-n!
Note that 2(1!)\!=2!,3(2!)\!=3!,4(3!)\!=4!, and so on
=2!\!+3!\!+4!\!+\cdots+n!\!+(n+1)!\!-1!\!-2!\!-3!\!-4!\!-\cdots-(n-1)!\!-n!
=(n+1)!\!-1!
=(n+1)!\!-1

Answer: (n+1)!\!-1

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About mvtrinh

Retired high school math teacher.
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