Let the statement P(n) has L.H.S. a recurrence relation tn+1 = 5 tn+4 , t = 4 and R.H.S. a general statement tn = 5n – 1.
Step I:
Put n = 1
L.H.S. = 4
R.H.S. = 51 – 1 = 4
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S. = t2 = 5t1+ 4 = 24
R.H.S. = t2 = 52 – 1 = 24
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 2.
Step II: Let us assume that P(n) is true for n = k.
∴ tk+1 = 5 tk+4 and tk = 5k – 1
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
Since tk+1 = 5 tk+4 and tk = 5k – 1 …..[From Step II]
tk+1 = 5 (5k – 1) + 4 = 5k+1 – 1
∴ 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, for all n ∈ N.