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Given that tn+1 = 5tn – 8, t1 = 3, prove by method of induction that tn = 5n+1 + 2

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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.

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