Prove that ∑(-1/2)^j = [2^(n+1) + (-1)^n]/3x2^n whenever n is a nonnegative integer

In this case, we will use Mathematical Induction.

PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive integers k.” Source: Discrete Mathematics and its Applications by Rosen.

∑(-1/2)^j = [2ⁿ⁺¹ + (-1)ⁿ]/(3×2ⁿ), the summation is from j=0 to n.

Let’s rewrite the formula,

1 + (-1/2) + (-1/2)² + … + (-1/2)ⁿ = [2ⁿ⁺¹ + (-1)ⁿ]/(3×2ⁿ)

Basis step:

P(0): 1= [2⁰⁺¹+(-1)⁰]/3×2⁰=3/3=1

Inductive hypothesis:

P(k): 1 + (-1/2) + (-1/2)² + … + (-1/2)^k = [2^k+1 + (-1)^k]/(3×2^k)

Inductive step:

P(k+1): 1 + (-1/2) + (-1/2)² + … + (-1/2)^k + (-1/2)^k+1= [2^k+2 + (-1)^k+1]/(3×2^k+1)

I will write every single step so you can fully understand the proof.

1 + (-1/2) + (-1/2)² + … + (-1/2)^k + (-1/2)^k+1, substituting P(k)

=[2^k+1 + (-1)^k]/(3×2^k) + (-1/2)^k+1

=[2^k+1 + (-1)^k]/(3×2^k) + (-1)^k+1/2^k+1

=[2^k+1 + (-1)^k]/(3×2^k) + (-1)^k+1/2×2^k

=[2(2^k+1 + (-1)^k) + 3(-1)^k+1]/6×2^k, if you didn’t get this step, think about how to solve a/2b+b/3b

=[2(2^k+1 + (-1)^k) + 3(-1)(-1)^k]/6×2^k

=[2(2^k+1 + (-1)^k) + 3(-1)(-1)^k]/6×2^k, 3(-1)=-3

=[2(2^k+1 + (-1)^k) – 3(-1)^k]/6×2^k

=[2×2^k+1 + 2(-1)^k – 3(-1)^k]/6×2^k, 2×2^k+1=2^k+2, 2-3=-1

=[2^k+2 – (-1)^k]/6×2^k

=[2^k+2 + (-1)(-1)^k]/3x2x2^k

=[2^k+2 + (-1)^k+1]/3×2^k+1

⧠

By the Principle of Mathematical Induction, ∑(-1/2)^j = [2ⁿ⁺¹ + (-1)ⁿ]/(3×2ⁿ) from j=1 to n, whenever n is a nonnegative integer.

Related exercises: