Question

Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + ... + (n × n!) = (n + 1)! − 1

Answer 1

P(n) is the statement

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

To prove for n = 1

L.H.S = 1! = 1

R.H.S = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1

L.H.S = R.H.S

⇒ P(1) is true

Assume that the given statement is true for n = k

(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true

To prove P(k + 1) is true

P(k + 1) = `"P"("k") + "t"_(("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

= (k + 1 + 1)! – 1

∴ P(k + 1) is true

⇒ P(k) is true,

So by the principle of mathematical induction

P(n) is true.