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