Suppose you wish to use mathematical induction to prove that: $$1 \cdot 1 ! + 2 \cdot 2 ! + 3 \cdot 3 ! + \cdots + n \cdot n ! = ( n + 1 ) ! - 1 \quad \text { for all } n \geq 1$$ (a) Write P(1). (b) Write P(5). (c) Write P(k). (d) Write P(k + 1). (e) Use mathematical induction to prove that P(n) is true for all n ≥ 1.

Question

Quizplus · Accepted Answer

The Answer is (a)

1 \cdot 1 ! = 2 ! - 1

(b)

1 \cdot 1 ! + 2 \cdot 2 ! + \cdots + 5 \cdot 5 ! = 6 ! - 1

(c)

1 \cdot 1 ! + 2 \cdot 2 ! + \cdots + k \cdot k ! = ( k + 1 ) ! - 1

(d)

1 \cdot 1 ! + 2 \cdot 2 ! + \cdots + ( k + 1 ) ( k + 1 ) ! = ( k + 2 ) ! - 1

(e)

P ( 1 )

is true since

1 \cdot 1 ! = 1

and

2 ! - 1 = 1 . \quad P ( k ) \rightarrow P ( k + 1 ) : 1 \cdot 1 ! + 2 \cdot 2 ! + \cdots + ( k + 1 ) ( k + 1 ) ! =

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

Suppose You Wish to Use Mathematical Induction to Prove That 1⋅1!+2⋅2!+3⋅3!+⋯+n⋅n!=(n+1)!−1 for all n≥11 \cdot 1 ! + 2 \cdot 2 ! + 3 \cdot 3 ! + \cdots + n \cdot n ! = ( n + 1 ) ! - 1 \quad \text { for all } n \geq 11⋅1!+2⋅2!+3⋅3!+⋯+n⋅n!=(n+1)!−1 for all n≥1

Suppose You Wish to Use Mathematical Induction to Prove That $1 \cdot 1 ! + 2 \cdot 2 ! + 3 \cdot 3 ! + \cdots + n \cdot n ! = ( n + 1 ) ! - 1 \quad \text { for all } n \geq 1$