# Unable to solve an Induction Problem......

Prove that all integers $$n>9$$ satisfy the inequality $$2^n>n^3$$ .

Note by Karthik Venkata
3 years, 6 months ago

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For any positive integer $$\displaystyle k$$, we have $$\displaystyle \dfrac{k+2}{k+1} < \dfrac{k+1}{k}$$

$$\displaystyle {2}^{\frac{1}{3}} > \dfrac{k+1}{k}$$ if $$\displaystyle k>9$$ by observation.

Thus, for all integers $$\displaystyle n>9$$, we have, cubing both sides $$\displaystyle 2k^3>(k+1)^3$$. It'll help us afterwards.

Canceling $$\displaystyle k^3$$ from both sides, we have $$\displaystyle k^3>3k^2+3k^1$$

By observation, for $$\displaystyle k=10$$, the base case, it is true that $$\displaystyle 2^k>k^3$$.

Let it be true for any $$\displaystyle k \geq 10$$. Then we have $$\displaystyle 2^k>k^3>3k^2+3k+1$$. Thus, $$\displaystyle 2^k+k^3>k^3+3k^2+3k+1=(k+1)^3$$.

Since $$\displaystyle 2^k>k^3$$, we can replace $$\displaystyle k^3$$ by $$\displaystyle 2^k$$.

Finally getting $$\displaystyle 2^{k+1}>(k+1)^3$$.

Hence, we have proved that if the claim is true for any integer $$\displaystyle k>9$$, it is also true for $$\displaystyle k+1$$.

Hence, the claim is true for all integers $$\displaystyle n>9$$

- 3 years, 6 months ago

What is the proof for your second step ? You said it is by observation. Is there any more complete way of proving the statement true for k+1, ie proving the inequality 2^(k+1)>(k+1)^3 from the induction hypothesis 2^k>k^3 for some k>9 ?

- 3 years, 6 months ago

That's what I did, but in the opposite order. Try using the facts I mentioned to get it in the right order.

- 3 years, 6 months ago

- 3 years, 5 months ago

I just gave RMO in 10th, and got 5 outta 6 correct, but still I didn't get selected. If you want a book, try "An Excursion in Mathematics".

- 3 years, 5 months ago

Oh my god, you mean to say that one needs to get a full score ? I am in class 9 presently, and after seeing previous year RMO papers, @-@ GONE MAD .Thanks for your suggestion, I will try that book.

- 3 years, 5 months ago

I never say that you need a full score. It's just that I solved 5 of 6, and didn't get selected. Maybe they didn't like my methods or solutions. And moreover, the papers of the recent years are a bit easier than those of the past years. All the best!

- 3 years, 5 months ago