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- Feb 14, 2012

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Prove that $3^n\ge(n+3)^3$ for any natural number $n\ge6$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,597

Prove that $3^n\ge(n+3)^3$ for any natural number $n\ge6$.

- Mar 31, 2013

- 1,283

This problem is same as

$3^{n-3} > = n^3$ or $3^n >= 27n^3$ for $n >=9$

To prove the same we use principle of mathematical induction

Base step

For n = 9 $LHS = 3^9 = 3^3 * 3^6 = 27 * 9^3$ so $3^n >= 27n^3$

So base step is true

Now $(\frac{n+1}{n})^3$ decreases as n increases and at n = 9 we have $(\frac{n+1}{n})^3= \frac{1000}{729}< 3$

So $(\frac{n+1}{n})^3< 3$ for all $n>=9$

Or $3 > (\frac{k+1}{k})^3\cdots(1)$ for all $k>=9$

Induction step

Let it be true for n = k $k >=9$

We need to prove it to be true for n = k+ 1

$3^k > = 27 k^3$

Multiplying by (1) on both sides

$3^{k+1} > = 27 * (\frac{k+1}{k})^3 * k^3 $

Or $3^{k+1} >= 27(k+1)^3$

So it is true for n = k+ 1

We have proved the induction step also

Hence proved