A factorial summation

Algebra Level 2

∑ n = 1 2018 n ! ⋅ n = ? \large \sum_{n=1}^{2018} n! \cdot n = ?

2019!-1 2019! 2018!+2019! 2019!+1

Santi Spadaro by Santi Spadaro , 39, Italy

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2 solutions

Santi Spadaro
Aug 9, 2018

Note that n ! ⋅ n = n ! ⋅ ( n + 1 − 1 ) = n ! ( n + 1 ) − n ! = ( n + 1 ) ! − n ! n! \cdot n= n! \cdot (n+1-1)= n!(n+1)-n!=(n+1)!-n! . Therefore the summation from the statement of the problem can be written as 2 ! − 1 ! + 3 ! − 2 ! + 4 ! − 3 ! + . . . + 2019 ! − 2018 ! 2!-1!+3!-2!+4!-3!+...+2019!-2018! , which after simplifying gives 2019 ! − 1 2019!-1 .

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Chew-Seong Cheong
Aug 12, 2018

S = ∑ n = 1 2018 n ! ⋅ n = ∑ n = 1 2018 n ! ( n + 1 − 1 ) = ∑ n = 1 2018 ( ( n + 1 ) ! − n ! ) = 2019 ! − 1 ! = 2019 ! − 1 \begin{aligned} S & = \sum_{n=1}^{2018} n! \cdot n = \sum_{n=1}^{2018} n! (n+1-1) = \sum_{n=1}^{2018} ((n+1)! - n!) = 2019! - 1! = \boxed{2019!-1} \end{aligned}

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(n+1)!=n! (n+1)=n! n+n!

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n!*n=(n+1)!-n!

sum(n!*n)=sum((n+1)!-n!)

sum((n+1)!-n!)=sum((n+1)!)-sum(n!)

sum((n+1)!)=sum(n!)+(n+1)!

Here the result must be 2019!

Olexander Danchuk - 5 months, 2 weeks ago

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Consider ∑ n = 1 3 ( n + 1 ) ! = 2 ! + 3 ! + 4 ! = 32 \displaystyle \sum_{n=1}^3 (n+1)! = 2! + 3! + 4! = 32 and ∑ n = 1 3 n ! + ( 3 + 1 ) ! = 1 ! + 2 ! + 3 ! + 4 ! = 33 ≠ 32 \displaystyle \sum_{n=1}^3 n! + (3+1)! = 1! + 2! + 3! + 4! = 33 \red{\ne 32} , ⟹ ∑ n = 1 N ( n + 1 ) ! ≠ ∑ n = 1 N n ! + ( N + 1 ) ! \displaystyle \implies \sum_{n=1}^N (n+1)! \red \ne \sum_{n=1}^N n! + (N+1)!

Chew-Seong Cheong - 5 months, 2 weeks ago

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