Big Summation I

Algebra Level 2

Evaluate: 1 × 1 ! + 2 × 2 ! + 3 × 3 ! + ⋯ + 2018 × 2018 ! + 2019 × 2019 ! 1\times 1! +2\times 2! +3\times 3!+\dots+2018\times 2018!+2019\times 2019!

Notation: ! ! denotes the factorial notation .

4040 ! 4040! 2021 ! 2021! 2020 ! + 1 2020! +1 2020 ! − 1 2020! -1 2020 ! 2020!

Hana Wehbi by Hana Wehbi , 51, USA

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3 solutions

Observe that n × n ! = ( n + 1 ) × n ! − n ! = ( n + 1 ) ! − n ! n\times n!=(n+1)\times n!-n!=(n+1)!-n! , then the summation becomes:

∑ k = 1 2019 n × n ! = ∑ k = 1 2019 [ ( n + 1 ) ! − n ! ] = 2020 ! − 1 ! = 2020 ! − 1 \quad\displaystyle\sum_{k=1}^{2019} n\times n! \\= \displaystyle\sum_{k=1}^{2019} [(n+1)!-n!] \\=2020!-1! \\=\boxed{2020!-1}

2 Helpful 1 Interesting 1 Brilliant 0 Confused
Hana Wehbi
Dec 30, 2019

We can add ( 1 = 1 ! ) (1 = 1!) to the summation:

( 1 + 1 ⋅ 1 ! ) + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ⋯ + 2018 ⋅ 2018 ! + 2019 ⋅ 2019 ! = (1+ 1\cdot1!) +2\cdot2! +3\cdot3!+\dots+2018\cdot2018!+2019\cdot2019! =

( 2 ! + 2 ⋅ 2 ! ) + 3 ⋅ 3 ! + ⋯ + 2018 ⋅ 2018 ! + 2019 ⋅ 2019 ! = (2! +2\cdot2!) +3\cdot3!+\dots+2018\cdot2018!+2019\cdot2019! =

( 3 ! + 3 ⋅ 3 ! ) + ⋯ + 2019 ⋅ 2019 ! = ( 3!+3\cdot3! ) + \dots + 2019\cdot2019! =

4 ! + 4 ⋅ 4 ! + ⋯ + 2019 ⋅ 2019 ! = ( 2019 ! + 2019 ⋅ 2019 ! ) = ( 2019 + 1 ) ! = 2020 ! 4!+4\cdot4! +\dots + 2019\cdot2019!= ( 2019!+2019\cdot2019!) = ( 2019 +1 )! = 2020!

⟹ 1 ⋅ 1 ! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ⋯ + 2018 ⋅ 2018 ! + 2019 ⋅ 2019 ! = 2020 ! − 1 \implies 1\cdot1! +2\cdot2! +3\cdot3!+\dots+2018\cdot2018!+2019\cdot2019! = 2020! - 1

3 Helpful 1 Interesting 0 Brilliant 0 Confused

Note that 1.1 ! = Γ ( 2.1 ) = 1.04649... 1.1! = \Gamma (2.1) = 1.04649... (see WolframAlpha ). That was why I said do not use decimal for multiplication sign.

Chew-Seong Cheong - 1 year, 5 months ago

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I won’t anymore. Thank you for letting me know.

Hana Wehbi - 1 year, 5 months ago

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I think you can edit your own solution here. Just use \cdot.

Chew-Seong Cheong - 1 year, 5 months ago

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@Chew-Seong Cheong – I did, let me know if there still anything l need to fix. Thank you.

Hana Wehbi - 1 year, 5 months ago
Chew-Seong Cheong
Dec 31, 2019

S = 1 ⋅ 1 ! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ⋯ + 2018 ⋅ 2018 ! + 2019 ⋅ 2019 = ( 2 − 1 ) ⋅ 1 ! + ( 3 − 1 ) ⋅ 2 ! + ( 4 − 1 ) ⋅ 3 ! + ⋯ + ( 2019 − 1 ) ⋅ 2018 ! + ( 2020 − 1 ) ⋅ 2019 = 2 ! + 3 ! + 4 ! + ⋯ + 2019 ! + 2020 ! − ( 1 ! + 2 ! + 3 ! + ⋯ + 2018 ! + 2019 ! ) = 2020 ! − 1 ! = 2020 ! − 1 \begin{aligned} S & = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + 2018 \cdot 2018! + 2019 \cdot 2019 \\ & = (2-1) \cdot 1! + (3-1) \cdot 2! + (4-1) \cdot 3! + \cdots + (2019-1) \cdot 2018! + (2020-1) \cdot 2019 \\ & = 2! + 3! + 4! + \cdots + 2019! + 2020! - (1! + 2!+3! + \cdots + 2018! + 2019!) \\ & = 2020! - 1! = \boxed{2020!-1} \end{aligned}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

@Hana Wehbi , don't use decimal point for multiplication sign but \cdot (center dot). Note the difference 1.1 1.1 1.1 and 1 \cdot 1 1 ⋅ 1 1\cdot 1 . Instead of explaining 1.1 = 1 × 1 1.1 = 1 \times 1 , just use \times in the problem to avoid confusions. Also ! is used as a math entity put it in LaTex. You answer options should be in LaTex because they involve math formulas. Of course, it will need extra keystrokes but it will make your problem look much more professional. It won't take long as you can see I have done all the editing for you.

Chew-Seong Cheong - 1 year, 5 months ago

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Thank you for all the remarks, l will definitely take them into consideration.

Hana Wehbi - 1 year, 5 months ago

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