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 \(\displaystyle 2^k\).

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 ?

@Karthik Venkata
–
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".

@Satvik Golechha
–
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.

@Karthik Venkata
–
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!

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## Comments

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TopNewestFor 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 \(\displaystyle n>9\)

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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 ?

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

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