- #1

- 11

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right side: [(k+1)+1]! - 1

left side: (k+1)! - 1 + (k+1) + (k+1)!

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- Mathematica
- Thread starter Mono182
- Start date

- #1

- 11

- 0

right side: [(k+1)+1]! - 1

left side: (k+1)! - 1 + (k+1) + (k+1)!

- #2

HallsofIvy

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- #3

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Using mathematical induction, prove that the following statements are true for n > 1, n = 1.

Q: 1 X 1! + 2 X 2! + ... + n X n! = (n + 1)! - 1

A: step 1: solve for n = 1

1 X 1! = (1 + 1)! - 1

1 = 1

step 2: assume true for n = k

therefore 1 X 1! + 2 X 2! + ... + k X k! = (k + 1)! - 1

step 3: sove for n = k + 1

1 X 1! + 2 X 2! + ... + k X k! + (k+1) X (k + 1)! = [(k+1) + 1)] - 1

since 1 X 1! + 2 X 2! + ... + k X k! = (k + 1)! - 1 , then

(k + 1)! - 1 + (k+1) X (k + 1)! = [(k+1) + 1)] - 1

now what i'm having trouble with is proving that the last statement is true

- #4

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Factor out (k+1)! (ignoring the -1)

- #5

VietDao29

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(k + 1)! - 1 + (k+1) X (k + 1)! = [(k+1) + 1)] - 1

now what i'm having trouble with is proving that the last statement is true

Yup, so far so good.

As

The LHS, and the RHS, both have "-1", so they cancel each other out, leaving you with:

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

You should also notice that:

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

Can you go from here? :)

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