Question

Prove by Mathematical Induction that 1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1​

Answer 1

Let p(n) = 1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1

Step 1:

First let us verify the result for n = 1

P(1) = 1! = (1 + 1)! – 1

P(1) = 1! = 2! – 1

P(1) = 1 = 2 – 1 = 1

∴ We have verified the result for n = 1.

Step 2:

Let us assume that the result is true for n = k

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

Step 3:

Let us prove the result for n = k + 1

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

P(k + 1) = P(k) + ((k + 1) × (k + 1)!)

P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!

= (k + 1)! + (k + 1) (k + 1)! – 1

= (k + 1)! (1 + k + 1) – 1

= (k + 1)! (k + 2) – 1

= (k + 2)! – 1

P(k + 1) = ((k + 1) + 1)! – 1

This implies P (k + 1) is true.

∴ Thus, we have proved the result for n = k + 1.

Hence by the principle of mathematical induction, the result is true for all natural numbers n.

(1 × 1!) + (2 × 2!) + (3 × 3!) + …………… + (n × n!) = (n + 1)! – 1

is true for all natural numbers n.