Is it true that ...? #4

Algebra Level 2

1 ( 1 ! ) + 2 ( 2 ! ) + 3 ( 3 ! ) + ⋯ + n ( n ! ) = ( n + 1 ) ! − 1 1(1!) + 2(2!) + 3(3!) + \cdots + n(n!) = (n+1)! - 1

Is the above true for all integers n ≥ 1 n \ge 1 ?

Yes No

Samuel Nascimento by Samuel Nascimento , 24, Brazil

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Jul 20, 2019

Let S n = 1 ( 1 ! ) + 2 ( 2 ! ) + 3 ( 3 ! ) + ⋯ + n ( n ! ) S_n = 1(1!) + 2(2!) + 3(3!) + \cdots + n(n!) , We can prove the claim that S n = ( n + 1 ) ! − 1 S_n = (n+1)!-1 is true by induction .

First note that S n = 1 ( 1 ! ) = 1 = 2 ! − 1 S_n = 1(1!) = 1 = 2!-1 . Therefore, the claim is true for n = 1 n=1 . Now, assuming that the claim is true for n n , then

S n + 1 = S n + ( n + 1 ) ( n + 1 ) ! = ( n + 1 ) ! − 1 + ( n + 1 ) ( n + 1 ) ! = ( n + 2 ) ( n + 1 ) ! − 1 = ( n + 2 ) ! − 1 = ( ( n + 1 ) + 1 ) ! − 1 \begin{aligned} S_{\color{#D61F06}n+1} & = S_n + (n+1)(n+1)! \\ & = (n+1)! - 1 + (n+1)(n+1)! \\ & = (n+2)(n+1)! - 1 \\ & = (n+2)! - 1 \\ & = ({\color{#D61F06}(n+1)}+1)! - 1 \end{aligned}

The claim is also true for n + 1 n+1 , therefore it is true for all n ≥ 1 n \ge 1 .

1 Helpful 0 Interesting 3 Brilliant 0 Confused

@Samuel Nascimento , I have changed the words of the question for you. You were asking if the member could prove it rather than if the equation is true. Someone else who knows that the equation is true may not be able to prove it. So is he or she going to answer "yes" or "no"?

Chew-Seong Cheong - 1 year, 10 months ago
Joseph Newton
Jul 21, 2019

For an arbitrary term k ( k ! ) k(k!) , we can write it as: k ( k ! ) = ( k + 1 ) ( k ! ) − k ! = ( k + 1 ) ! − k ! \begin{aligned}k(k!)&=(k+1)(k!)-k!\\&=(k+1)!-k!\end{aligned} This gives us a telescoping sum: 1 ( 1 ! ) + 2 ( 2 ! ) + 3 ( 3 ! ) + ⋯ + n ( n ! ) = ( 2 ! − 1 ! ) + ( 3 ! − 2 ! ) + ( 4 ! − 3 ! ) + ⋯ + ( ( n + 1 ) ! − n ! ) = − 1 ! + 2 ! − 2 ! + 3 ! − 3 ! + 4 ! + … − n ! + ( n + 1 ) ! = − 1 ! + ( n + 1 ) ! = ( n + 1 ) ! − 1 \begin{aligned}1(1!)+2(2!)+3(3!)+\dots+n(n!)&=(2!-1!)+(3!-2!)+(4!-3!)+\dots+((n+1)!-n!)\\ &=-1!\color{#D61F06}+2!-2!\color{#3D99F6}+3!-3!\color{#20A900}+4!+\color{#333333}\dots\color{#624F41}-n!\color{#333333}+(n+1)!\\ &=-1!+(n+1)!\\ &=(n+1)!-1\end{aligned}

0 Helpful 0 Interesting 3 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...