Question

Show that 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n by the Principle of Mathematical Induction.​

Answer 1

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$\Large\fbox{\color{purple}{QUESTION}}$

Show that 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n by the Principle of Mathematical Induction.

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

➡️Assume that P(n) be the given statement, that is

➡️P(n) : 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n.

➡️It is noted that P (1) is true, since

➡️P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1.

➡️Let P(n) is true for some natural number k,

➡️It means that

➡️P(k) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! = (k + 1)! – 1

➡️Inorder to prove P (k + 1) is true, we have

➡️P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! + (k + 1) × (k + 1)!

➡️= (k + 1)! – 1 + (k + 1)! × (k + 1)

➡️Now, simplify the above form, we get

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

➡️Therefore, P (k + 1) is true, whenever P (k) is true.

➡️Hence, P(n) is true for all natural number n is proved using the Principle of Mathematical Induction.

$\Large\mathcal\green{FOLLOW \: ME}$

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Answer 2

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Assume that P(n) be the given statement, that is

P(n) : 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 for all natural numbers n.

It is noted that P (1) is true, since

P (1) : 1 × 1! = 1 = 2 – 1 = 2! – 1.

Let P(n) is true for some natural number k,

It means that

P(k) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! = (k + 1)! – 1

Inorder to prove P (k + 1) is true, we have

P (k + 1) : 1 × 1! + 2 × 2! + 3 × 3! + … + k × k! + (k + 1) × (k + 1)!

= (k + 1)! – 1 + (k + 1)! × (k + 1)

Now, simplify the above form, we get

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

Therefore, P (k + 1) is true, whenever P (k) is true.

Hence, P(n) is true for all natural number n is proved using the Principle of Mathematical Induction.