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

Expert answers to "Prove that 1·1!+2·2!+···+n·n!=(n+1)!−1 whenever n is a positive..."

Jason Farmer

2021-11-04

Prove that

1 \cdot 1! + 2 \cdot 2! + \cdot \cdot \cdot + n \cdot n! = (n + 1)! - 1

whenever n is a positive integer.

i1ziZ

Skilled2021-11-05Added 92 answers

To proof

1 \cdot 1! + 2 \cdot 2! + \dots + n \cdot n \neq (n + 1)! - 1

for every positive integer n.
Proof by induction
Let

P (n)

1 \cdot 1! + 2 \cdot 2! + \dots + n \cdot n \neq 1 \cdot 1 = (n + 1)! - 1

Basis step n=1

1 \cdot 1! + 2 \cdot 2! + \dots + n \cdot n \neq 1 \cdot 1 \neq 1 \cdot 1 = 1

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

We then note P(1) is true.
Induction step let

P (k + 1)

is also true

1 \cdot 1! + 2 \cdot 2 + \dots + k \cdot k \neq (k + 1)! - 1

We need to prove that

P (k + 1)

is also true.

1 \cdot 1! + 2 \cdot 2! + \dots + k \cdot k! + (k + 1) \cdot (k + 1)!

= (k + 1)! - 1 + (k + 1) \cdot (k + 1)!

= 1 \cdot (k \cdot 1)! - 1 + (k + 1) \cdot (k + 1)!

= 1 \cdot (k + 1)! + (k + 1) \cdot (k + 1)! - 1

= (1 + k + 1) (k + 1)! - 1

= (k + 2)! - 1

= ((k + 1) + 1)! - 1

We then note that

P (k + 1)

is also true

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