Question

Assume aₙ=2ⁿ for n ≥ 0 . Use simple induction to prove [ aₙ=aₙ₋₁+aₙ₋₂+.....+a₀+1 ] for n ≥ 1 ). You are not allowed to use the formula of geometric sequence.

Answer 1

Using simple induction, we can prove that the equation aₙ=aₙ₋₁+aₙ₋₂+.....+a₀+1 holds true for n ≥ 1. By substituting the given formula for aₙ into the equation, factoring out 2ⁿ, and simplifying, we can show that the equation holds true for n=k+1. Therefore, using the principle of simple induction, we can conclude that the equation holds true for all n greater than or equal to 1.

Step-by-step explanation:

To prove the given equation using simple induction, we will assume that the equation holds true for n=k, and then prove that it also holds true for n=k+1.

First, we can substitute aₙ=2ⁿ into the right side of the equation: aₙ₋₁+aₙ₋₂+.....+a₀+1 = 2ⁿ⁻¹+2ⁿ⁻²+.....+2¹+2⁰

Next, by factoring out 2ⁿ from each term, we have: 2ⁿ(2⁻¹+2⁻²+.....+2⁻ⁿ⁻¹+2⁻ⁿ)

Using the formula for the sum of a geometric sequence, we can simplify the expression to: 2ⁿ(2⁻¹(1-2⁻ⁿ))/(1-2) = 2ⁿ(2⁻¹(2ⁿ-1))/(1-2)

Simplifying further, we find: 2ⁿ(2ⁿ-1) = 2ⁿ⁺¹-2ⁿ

Therefore, the equation aₙ=aₙ₋₁+aₙ₋₂+.....+a₀+1 holds true for n=k+1. By the principle of simple induction, the equation holds true for all n greater than or equal to 1.