Prove that:

1(1!) + 2(2!) + ...... + n(n!) = (n+1)! -1

Here is what I havedone so far but I am stuckk?

(n+1)! -1 +((n+1) (n+1)!)

= (n+1)![1 + (n+1)] -1

I cannot ge it to factor own to what it is suppose to be, please help!

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- February 15th 2007, 02:02 PMluckyc1423induction
Prove that:

1(1!) + 2(2!) + ...... + n(n!) = (n+1)! -1

Here is what I havedone so far but I am stuckk?

(n+1)! -1 +((n+1) (n+1)!)

= (n+1)![1 + (n+1)] -1

I cannot ge it to factor own to what it is suppose to be, please help! - February 15th 2007, 04:10 PMPlato
Have you even started? Are you even on the path?

Is it true for n=1? - February 15th 2007, 07:33 PMluckyc1423
Yes its true for all those numbers, I have already seen that by my notes...I cannot factor it anymore for some reason...The assignment is due by 8AM

- February 15th 2007, 08:38 PMSoroban
Hello, luckyc1423!

Quote:

Prove that:

. . 1(1!) + 2(2!) + 3(3!) + ... + n(n!) .= .(n+1)! -1

You should write out the steps . . . clearly.

Verify S(1): .1(1!) .= .2! - 1 . . . true

Assume S(k): .1(1!) + 2(2!) + 3(3!) + ... + k(k!) .= .(k+1)! - 1

We will try to prove that S(k+1) is true.

. . You might write it out (on the side) just for reference.

Add (k+1)(k+1)! to both sides:

. . 1(1!) + 2(2!) + 3(3!) + ... + (k+1)(k+1)! .= .(k+1)! - 1 + (k+1)(k+1)!

The left side is the left side of S(k + 1).

The right side is: .(k+1)! + (k+1)(k+1)! - 1

. . . . . . . Factor: .(k+1)![1 + (k + 1)] - 1

. . And we have: . (k+1)!(k+2) - 1 .= .(k+2)! - 1

This is the right side of S(k+1). .The inductive proof is complete.