Question

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

Answer 1

Attempt: We will use induction
First for the basis case, if we let n = 1 then

1 · 1! + 2 · 2!+· · ·+1 · 1! = (1 + 1)! − 1
1 · 1! + 2 · 2!+· · ·+1 · 1! = (2)! − 1
1 · 1! + 2 · 2!+· · ·+1 · 1! = (2)! − 1
(don't know why I have that massive left side still there when it should only be)
1 · 1! = (2)! − 1
1 · 1 = (2) − 1
1=1
so the basis case is true.

Answer 2

now for the k case, if we let n = k

1 · 1! + 2 · 2!+· · ·+k · k! = (k + 1)! − 1

and then for the k+1 case
1 · 1! + 2 · 2!+· · ·+k+1 · k+1! = (k+1 + 1)! − 1
1 · 1! + 2 · 2!+· · ·+k+1 · k+1! = (k+2)! − 1

so the right hand side of the equation is my goal line so I have to manipulate the left to get (k+2)! -1 but I keep on missing a step and everything falls apart.

Answer 3

so first I go back to the k case and then just add the k+1
1 · 1! + 2 · 2!+· · ·+k · k! = (k + 1)! − 1
1 · 1! + 2 · 2!+· · ·+k · k! + k+1 ·k+1! = (k + 2)! − 1

and since 1 · 1! + 2 · 2!+· · ·+k · k! = (k + 1)! − 1
I have to substitute
(k + 1)! − 1+ k+1 ·k+1! = (k + 2)! − 1

since there is a (k+1)! in common, then maybe I could factor , but then that leads

(k+1)! (1-1+k+1) = (k+2)!-1
then I'm stuck...

Answer 4

Woops :) I see your mistake.
You factored a (k+1)! out of too many terms.
Notice that only 2 of your 3 terms have a (k+1)!, not the -1 in the middle though.

Answer 5

yeah....that's where I screwed up D:

Answer 6

\[\large\rm \color{royalblue}{(k+1)!}-1+(k+1)\color{royalblue}{(k+1)!}\quad=\quad \color{royalblue}{(k+1)!}\left[1+(k+1)\right]-1\]

Answer 7

but what happens next
combine like terms in the []?
(k+1)![1+(k+1)]-1
(k+1)![1+k+1]-1
(k+1)![k+2]-1 ????????????

Answer 8

Good :) It's not jumping out at you from there?

Answer 9

Example: \(\large\rm 4!\cdot5=5!\)

Answer 10

idk I've been sick for a week, so my mind is blown already... what's jumping out?

Answer 11

\[\large\rm 6!\cdot7=7!\]See how these are consecutive values?
Try to apply that to your next step,\[\large\rm (k+1)!\cdot(k+2)\quad=\quad ?\]

Answer 12

\[\large\rm (k+1)!\cdot(k+2)\quad=(k+2)! \]
like that ?

Answer 13

Yes :) because k+2 is the next integer value after k+1.\[\large\rm 6!\cdot7=1\cdot2\cdot\cdot\cdot6\cdot7=7!\]
So,\[\large\rm (k+1)!\cdot(k+2)\quad=1\cdot2\cdot\cdot\cdot(k+1)\cdot(k+2)\quad=(k+2)!\]

Answer 14

wow that part must have slipped out of my mind, but by doing that we do have LHS = RHS so induction holds :)

Answer 15

Woooo nice job \c:/

Answer 16

yay :D