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.

  • $\begingroup$ 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

Your inequality, after dividing by $5^n$ is equivalent to




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


which is the desired inequality.

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


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


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


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







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


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