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Question
Prove by method of induction, for all n ∈ N:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1
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Solution
The statement P(n) has L.H.S. a recurrence relation tn+1 = 5tn + 4, t1 = 4 and R.H.S. a general statement tn = 5n – 1
Step I:
To prove P(1) is true.
L.H.S. = 4, R.H.S. = 51 – 1 = 4
∴ P(1) is true.
For n = 2, L.H.S. = t2 = 5t1 + 4 = 24
R.H.S. = t2 = 52 – 1 = 24
∴ P(2) is also true.
Step II:
Assume P(k) is true.
i.e. tk+1 = 5tk + 4 and tk = 5k – 1
Step III:
To prove P(k + 1) is true.
i.e. to prove tk+1 = 5k+1 – 1
Since tk+1 = 5tk + 4 and tk = 5k – 1 .........(From Step II)
∴ tk+1 = 5(5k – 1) + 4 = 5k+1 – 1
∴ P(k + 1) is true.
From all the steps above,
P(n) : tn = 5n – 1 is true for all
n ∈ N where tn+1 = 5tn + 4 , t1 = 4
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