Prove that 2−2·7+2·7^2 −···+2(−7)^n =(1− (−7)^(n+1))/4 whenever n is a nonnegative integer

To make this proof, we will use the principle of 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.

Basis step:

P(0): 2 = (1- (-7)⁰⁺¹)/4

(1- (-7)⁰⁺¹)/4 = (1 – (-7)/4 = 8/4 = 2.

Notice that in this case, the basis step is P(0) and P(1) because the formula is valid for all non-negative integers, and 0 is a nonnegative integer.

Inductive hypothesis:

P(k): 2−2·7+2·7² −···+2(−7)^k =(1− (−7)^k+1)/4

Inductive step:

P(k+1): 2−2·7+2·7² −···+2(−7)^k + 2(-7)^k+1 =(1− (−7)^k+²)/4

2−2·7+2·7² −···+2(−7)^k + 2(-7)^k+1 Substituting P(k) we get,

=(1− (−7)^k+1)/4+ 2(-7)^k+1

=[1− (−7)^k+1+ 8(-7)^k+1]/4 if you don’t get this steep, think that 1/4 +3 = (1+3×4)/4

=[1+ 7(-7)^k+1]/4 (-1a +8a=7a)

=[1 – (-7)(-7)^k+1]/4, using the fact that axaⁿ=aⁿ⁺¹ we get,

= (1- (-7)^k+2)/4

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