Prove that 1·1!+2·2!+···+n·n!=(n+1)!-1 whenever n is a positive integer. Mathematical Induction.

Question

This is a problem that is supposed to use mathematical induction.

I know how to create the basis case, and create the assumption using "k" but as soon as you are supposed to prove for "k+1" i am lost...please help me!

Answer 1

For (k+1), you want to show that [1•1!+2•2!+...+k•k!]+(k+1)•(k+1)! = (k+1+1)!-1 = (k+2)!-1

 

[1•1!+2•2!+...+k•k!]+(k+1)•(k+1)!     *write out the equation for k+1, then you can use the 'k' assumption to condense that part of the equation into

[(k+1)!-1]+(k+1)•(k+1)!

 

 

Then you can rearrange that to look like

 

(k+1)!+(k+1)!•(k+1)-1

Factor out (k+1)! to get

(k+1)!•(1+(k+1))-1

=(k+1)!(k+2)-1   *(k+2)(k+1)! is equal to (k+2)!*

=(k+2)!-1

QED