Prove, by induction, that $5^n + 5 < 5^{n+1}$ for all $n\in\Bbb N$.

Question

Attempt:

If $n = 1$, then $5^1 + 5 < 5^{2}$ => $10 < 25$ which is a true statement so the base case holds.

Assume $5^k + 5 < 5^{k+1}$ is true for some $k\in\Bbb N$.

How to prove it for $n = k+1$?

Write the statement for $n=k+1$ and see if you can derive it from the case $n=k$. — Carlos Eugenio Thompson Pinzón, Dec 05 '13 at 22:40

score 1 · Answer 1 · answered Dec 05 '13 at 22:40

1

$$5^{k+1} + 5 < 5^{k+1} + 25 = 5(5^k + 5) < 5 \cdot 5^{k+1} = 5^{k+2}$$

answered Dec 05 '13 at 22:40

Arthur

ILoveMath · Answer 2 · 2013-12-06T08:07:48.323

1

Notice for $m,n$ positive integers and $x,y \geq 2$, we have

$$ x^m + y^n \leq x^my^n $$

Now, put $x = y = 5$ and $m = 1 $

$$ \therefore 5^1 + 5^n \leq 5^1 5 ^n \implies 5^n + 5 \leq 5^{n+1} $$

as desired.

edited Dec 06 '13 at 08:07

answered Dec 05 '13 at 23:53

ILoveMath

score -1 · Answer 3 · answered Dec 06 '13 at 08:20

-1

$$ \color{#0000ff}{\large 5^{n + 1} + 5} = 5\left(5^{n} + 5\right) - 20 < 5\times 5^{n + 1} - 20 \color{#0000ff}{\large < 5^{n + 2}} $$

answered Dec 06 '13 at 08:20

Felix Marin

3 Answers3