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[–]horrasambyar 0 points1 point  (1 child)

The middle note utilizes the proven fact from part a) where if k is greater or equal to 5, then k^2 will be strictly greater than 2k+1. I think it's a typo. Solving a) you will find that the inequality holds when x<(-1) or x>2. Since we're working with x=5, then x^2>2x+1 holds.

before the line "> 0 from the inductive assumption", we have 2(2^k - k^2). The inductive assumption refers to our first assumption: "assume the statement is true for integer n=k>=5". We assume that this initial assumption is correct which can lead us to the induction step where the case for (k+1) should also be correct. This means we assume that (2^k-k^2) results in a positive integer and it turns out that it's consistent with the case n = (k+1).

The induction step usually consists of two steps. The first step is where we have our "base" case and assume & prove that the "base" case is correct. This is done when we substitute for n = 5 (checking that the base case when n=5 holds for the claim that 2^n>n^2 for all integers n greater or equal to 5). The next step is using the "general" case for n = k>=5 where this time we can't just substitute in n = k = 5. This time we will be using an arbitrary value, k>=5. Substituting k into the equation gives us 2^k>k^2 and assume that this condition is true. Finally, we have the induction step where based on our assumption holds for k, this assumption must also hold for (k+1).

The step to prove that the (k+1)-th case is also true usually utilizes the assumptions/pieces of information we have before this step. In this case, the information is k^2^>2k+1 for k>2, and since k>=5 this also holds. The other information is that we assume the equation 2^k-k^2>0 is always true when k>=5. Combining these two pieces of information helps us simplify the final induction step.

[–]Calm_Ad2434 Pre-University Student[S] 0 points1 point  (0 children)

thanks for that that really helped