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 = [2k+1 + (-1)k]/(3×2k)
Inductive step:
P(k+1): 1 + (-1/2) + (-1/2)2 + … + (-1/2)k + (-1/2)k+1= [2k+2 + (-1)k+1]/(3×2k+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)
=[2k+1 + (-1)k]/(3×2k) + (-1/2)k+1
=[2k+1 + (-1)k]/(3×2k) + (-1)k+1/2k+1
=[2k+1 + (-1)k]/(3×2k) + (-1)k+1/2×2k
=[2(2k+1 + (-1)k) + 3(-1)k+1]/6×2k, if you didn’t get this step, think about how to solve a/2b+b/3b
=[2(2k+1 + (-1)k) + 3(-1)(-1)k]/6×2k
=[2(2k+1 + (-1)k) + 3(-1)(-1)k]/6×2k, 3(-1)=-3
=[2(2k+1 + (-1)k) – 3(-1)k]/6×2k
=[2×2k+1 + 2(-1)k – 3(-1)k]/6×2k, 2×2k+1=2k+2, 2-3=-1
=[2k+2 – (-1)k]/6×2k
=[2k+2 + (-1)(-1)k]/3x2x2k
=[2k+2 + (-1)k+1]/3×2k+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:
- a) Find a formula for 1/(1×2) + 1/(2×3) + 1/n(n+1) by examining the values of this expression for small values of n. b)Prove the formula you conjectured in part (a)
- Prove that 2−2·7+2·7^2 −···+2(−7)^n =(1− (−7)^(n+1))/4 whenever n is a nonnegative integer
- Let P(n) be the statement that 1^3 +2^3 +···+n^3 = (n(n + 1)/2)^2 for the positive integer n
