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

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- Dec 12th 2009, 04:39 PMleinadwerdnai struggle with proofs by induction
Prove by induction that for all integers n greater than or equal to 2, 2^n+1 < 3^n

- Dec 12th 2009, 05:36 PMcraig
This is how I would do it:

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

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

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

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

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

Dividing both sides by 2

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

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

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

QED - Dec 15th 2009, 03:07 PMleinadwerdna
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