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I'm trying to prove that $5^n-3^n>5^{n-1}$

I tried using mathematical induction and got stuck at the induction step.

First, I started by rearranging the inequality as: $4 \times5^n>5\times3^n$

  • Try $n=1$: $$20>15$$ Therefore true for $n=1$
  • Assume true for $n=k$: $$4 \times5^k>5\times3^k$$
  • Examine case $n=k+1$: $$4\times5^{k+1}>5\times3^{k+1}$$

I'm not really sure where to go from here. Any help would be appreciated.

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  • $\begingroup$ The lhs grows by a factor of $5$, the rhs only by a factor of $3$. Use $4\cdot 5^{k+1}=5\cdot 4\cdot 5^k$. $\endgroup$ – Hagen von Eitzen Oct 16 '16 at 19:50
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Your inequality, after dividing by $5^n$ is equivalent to

$1-(\frac{3}{5})^n>\frac{1}{5}$

or

$(\frac{3}{5})^n<\frac{4}{5}$

for $n=1$, it is true.

Now, let $n\geq 1$ such that

$(\frac{3}{5})^n<\frac{4}{5}$ (induction hypothesis).

as $\frac{3}{5}<1$

if we multiply by $(\frac{3}{5})^n$, we will get

$(\frac{3}{5})^{n+1}<(\frac{3}{5})^n<\frac{4}{5}$

which is the desired inequality.

we conclude that the inequality is satsfied for all integer $n\geq 1$.

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First, show that this is true for $n=1$:

$5^1-3^1>5^{1-1}$

Second, assume that this is true for $n$:

$5^n-3^n>5^{n-1}$

Third, prove that this is true for $n+1$:

$5^{n+1}-3^{n+1}=$

$5\cdot5^n-3\cdot3^n>$

$5\cdot5^n-5\cdot3^n=$

$5\cdot(\color\red{5^n-3^n})>$

$5\cdot\color\red{5^{n-1}}=$

$5^n$


Please note that the assumption is used only in the part marked red.

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