Question

: Let a1, a2, a3, . . . be the sequence of integers defined by a₁ = 1 and defined for n ≥ 2 by the recurrence relation a_n = 3a_n−1 + 1. Using the Principle of Mathematical Induction, prove for all integers n ≥ 1 that a_n = (3 ⁿ − 1) /2 .

Answer 1

Base-Case (n=2)

Using recurrence relation, a2 3a1 +1 3(1) 14

(32 - 1) )-- Using formula, a2

Hence the base case is satisfied

Assumption Step (n=k)

Let us assume that the given thing holds true for n=k, so we can write

$a_k = \frac{3^k - 1}{2}$

Inductive Step (n=k+1)

Now, we need to prove that the given things hold true for (n=k+1)

3+1-1 3-1 +1

which is similar to the formula

$a_{k+1} = \frac{3^{k+1} - 1}{2}$

Hence the given thing holds true for (n=k+1)

Therefore, using the principle of mathematical induction, the equation is valid for all integers n>=1