i struggle with proofs by induction

**leinadwerdna** · December 12th 2009, 04:39 PM

Prove by induction that for all integers n greater than or equal to 2, 2^n+1 < 3^n

**craig** · December 12th 2009, 05:36 PM

Originally Posted by leinadwerdna

Prove by induction that for all integers n greater than or equal to 2, 2^n+1 < 3^n

This is how I would do it:

For $n=2$ , $2^3 < 3^2$ , $8<9$ , true for $n=2$

Assume for $n=k$ , this gives you $2^{k+1} < 3^k$

Now the induction step, prove for $n=k+1$

$2^{k+2} < 3^{k+1}$

$2 \cdot 2^{k+1} < 3 \cdot 3^{k}$

Dividing both sides by 2

$2^{k+1} < \frac{3}{2} \cdot 3^{k}$

Therefore if $2^{k+1} < 3^k$ , then $2^{k+1}$ is definitely $< \frac{3}{2} \cdot 3^{k}$ , true for all $n=k+1$ if true for $n=k$

True for $n=2$ , therefore true for all integers greater than or equal to 2.

QED

**leinadwerdna** · December 15th 2009, 03:07 PM

for the induction step couldnt you also do with 2^(k+1)<3^k multiply both sides by 2 of that so you get 4(2^k) < 2(3^k). And since we are trying to prove 4(2^k)<3(3^k) this obviously proves it since 2 is less than 3

does this make any sense

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