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Given that tn+1= 5 tn+4 , t1 = 4, prove by method of induction that tn = 5n – 1

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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.

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