recursive relations

Question

how does a_n = 2a_n-1 + 3, a₁ = 3 become 3(2ⁿ -1)

Answer 1

Using this recursive formula, lets start looking for a pattern

a_n = 2a_n-1 + 3

a₁ = 2a₀ + 3

a₂ = 2a₁ + 3 = 2(2a₀ + 3) + 3 = 2*2*a₀ + 3 + 2*3 = 4a₀ + 9

a₃ = 2a₂ + 3 = 2(4a0 + 9) + 3 = 2*2*2*a₀ + 3 + 2*3 + 2*2*3 = 8a₀ + 21

...

Therefore the nth term would have

a_n = 2ⁿa₀ + 3 + 2*3 + 2*2*3 + ... (2*2*2 n-1 terms)*3

a_n = 2ⁿa₀ + 3 * (1 + 2 + 2*2 + 2*2*2 + ... + (2*2*2 n-1 terms))

The last term is nothing more than the sum for 2ⁿ-1...

a_n = 2ⁿa₀ + 3*(2ⁿ-1)

We solve for a₀ using the fact that a₁ = 3. According to first recursion, a₁ = 2a₀ + 3 which yields a₀ = 0.

Finally we arrive at a_n = 3*(2ⁿ-1)