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Question
By the principle of mathematical induction, prove the following:
1 + 4 + 7 + ……. + (3n – 2) = `("n"(3"n" - 1))/2` for all n ∈ N.
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Solution
Let P(n) : 1 + 4 + 7 + ……. + (3n – 2) = `("n"(3"n" - 1))/2`
Put n = 1,
LHS = 1
RHS = `(1(3 - 1))/2 = 1`
∴ P(1) is true.
Assume P(k) is true for n = k
P(k): 1 + 4 + 7 + ……. + (3k – 2) = `(k(3k - 1))/2`
To prove P(k + 1) is true, i.e., to prove
1 + 4 + 7 + ……. + (3k – 2) + (3(k + 1) – 2) = `((k + 1)(3(k + 1) - 1))/2`
1 + 4 + 7 + ……. + (3k – 2) + (3k + 3 – 2) = `((k + 1)(3k + 2))/2`
1 + 4 + 7 + …… + (3k + 1) = `((k + 1)(3k + 2))/2`
1 + 4 + 7 + …… + (3k – 2) + (3k + 1) = `(k(3k - 1))/2 + (3k + 1)`
`= (k(3k - 1) + 2(3k + 1))/2`
`= (3k^2 - k + 6k + 2)/2`
`= (3k^2 + 5k + 2)/2`
`= ((k + 1)(3k + 2))/2`
∴ P(k + 1) is true whenever P(k) is true.
∴ By the Principle of Mathematical Induction, P(n) is true for all n ∈ N.
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