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प्रश्न
Answer the following:
Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2
उत्तर
The statement P(n) has L.H.S. a recurrence relation tn+1 = 5tn – 8, t1 = 3 and R.H.S. a general statement tn = 5n–1 + 2
Step I:
To prove P(1) is true.
L.H.S. = 3, R.H.S. = 51–1 + 2
= 1 + 2
= 3
So P(1) is true
∴ For n = 2, L.H.S = t2 = 5t1 – 8 = 5(3) – 8 = 7
R.H.S. = t2 = 52–1 + 2
= 5 + 7
= 2
∴ P(2) is also true
Step II:
Assume P(k) is true
i.e. tk+1 = 5tk – 8 and tk = 5k–1 + 2
Step III:
To prove P(k + 1) is true
i.e. to prove tk+1 = 5k+1–1 + 2 = 5k + 2
Since tk+1 = 5tk – 8 and tk = 5k–1 + 2 ......(from step II)
∴ tk+1 = 5(5k–1 + 2) – 8 = 5k + 2
∴ P(k + 1) is true.
From all the steps above,
P(n) : tn = 5n–1 + 2 is true for all
n ∈ N where tn+1 = 5tn – 8, t1 = 3.
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