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

Question

Theorem

Solution

Let 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.

Note that P(1) is true.

Since P(1): 1 × 1! = 1

= 2 – 1

= 2! – 1.

Assume that P(n) is true for some natural number k.

i.e., P(k) : 1 × 1! + 2 × 2! + 3 × 3! + ... + k × k! = (k + 1)! – 1

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)

= (k + 1 + 1) (k + 1)! – 1

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

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

Therefore, by the Principle of Mathematical Induction, P(n) is true for all natural number n.

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Principle of Mathematical Induction

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Chapter 4: Principle of Mathematical Induction - Solved Examples [Page 67]

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NCERT Exemplar Mathematics Exemplar [English] Class 11

Chapter 4 Principle of Mathematical Induction
Solved Examples | Q 9 | Page 67

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