Let 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:
Put n = 1
L.H.S. = 3
R.H.S. = 51-1 + 2 = 1 + 2 = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S = t2 = 5t1 – 8 = 5(3) – 8 = 7
R.H.S. = t2 = 52-1 + 2 = 5 + 2 = 7
∴ L.H.S. = R.H.S.
∴ P(n) is term for n = 2.
Step II:
Let us assume that P(n) is true for n = k.
∴ tk+1 = 5tk – 8 and t = 5k-1 + 2
Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
tk+1 = 5k+1-1 + 2 = 5k + 2
tk+1 = 5tk – 8 and tk = 5k-1 + 2 ……[From Step II]
∴ tk+1 = 5(5k-1 + 2) – 8 = 5k + 2
∴ P(n) is true for n = k + 1.
Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ tn = 5n-1 + 2, for all n ∈ N.