Here is what I have so far


For $n = 0 (1-0.3)^0 \geq 1-0.3(0)$ checks

For $n = 1 (1-0.3)^1 \geq 1-0.3(k$) checks

I.H. $(1-0.3)^k \geq 1-0.3(k)$ for all k in the set of positive integers (1)

We want to prove that $(1-0.3)^{k+1} \geq 1-0.3(k+1)$ (2)

To relate (1) to (2) we have to add $(1-0.3)^{k+1}$ to both sides of (1).

$(1-0.3)^k+(1-0.3)^{k+1} \geq 1-0.3(k) + (1-0.3)^{k+1}$.

Here is where I am stuck. I need help as to where to go from here.

  • $\begingroup$ Hint : Try binomial expansion of (1-0.3)^k+1 in equation 2 . $\endgroup$ – Lakshya Gupta Apr 24 '17 at 23:52
  • $\begingroup$ "To relate (1) to (2) we have to add (1−0.3)k+1 to both sides of (1)." Why? $\endgroup$ – fleablood Apr 25 '17 at 0:13

I don't see why you have to add $(1-0.3)^{k+1}$.

\begin{align} (1-0.3)^{k+1} &= (1-0.3)(1-0.3)^k \\ &\geq (1-0.3) (1-0.3k) \\ &= 1-0.3-0.3k+0.3^2k \\ &=1-0.3(k+1)+0.3^2k \\ &\geq 1-0.3(k+1) \end{align}

  • $\begingroup$ I was told that when solving this problem. You have to add (1-0.3)^k+1 to both sides of the induction hypotheses. $\endgroup$ – Phillip Mc Apr 25 '17 at 0:02
  • $\begingroup$ Did it occur to you to ask the person who told you that why? Seeing as we didn't tell you any such thing, I don't see that we have to pay any attention to it. $\endgroup$ – fleablood Apr 25 '17 at 0:11

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