Melanie W. answered • 03/12/15

Physics and Math tutor to help you succeed

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**