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I am trying to prove for all natural $n$ that: $$5^n + 5 < 5^{n+1}$$

I did the basic step with $n=1$ and inequality holds, I am now at the induction step: $$5^{k+1} + 5 < 5^{k+2}$$

and I have no idea how to proceed from here. Can someone give me a clue?

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The second hint: for positive $x,y:$ if $x > y + 25$ then $x > x + 5.$ –  user2468 Jul 15 '12 at 17:46

4 Answers 4

At the induction step, assuming the inequality is true for $n=k$, we have $$5^{k} + 5 < 5^{k+1}$$ Multiplying the above by $5$ and make use of the fact the $5^{k+1} + 5 < 5^{k+1} + 25$ to conclude what you want.

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This is correct and the simplest inductive solution but I think it's better to just give hints on a question tagged homework. –  D Coetzee Jul 15 '12 at 18:49
    
@DerrickCoetzee What I wrote was meant to be a hint. –  user17762 Jul 15 '12 at 19:09

For any $\,n\in\Bbb N=\{1,2,3,...\}\,$:

$$5+5^n<5^{n+1}\stackrel{\text{div. by}\,5}\Longleftrightarrow 1+5^{n-1}<5^n\Longleftrightarrow 1<4\cdot 5^{n-1}$$ And since $\,4\cdot 5^{n-1}\,$ is the product of two natural numbers, one of them greater than $\,1\,$, we're done.

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As the OP wrote, it is a homework problem, to be proved using induction, so skipping the induction is probably not going to help the OP. One way is to inductively prove: a product of integers $> 1$ is $> 1$, a special case of the method I use, which I exploit heavily in my posts on telescopy (see the links in my answer). –  Bill Dubuque Jul 15 '12 at 18:43

Hint $\rm\ 5\! +\! 5^n < 5^{n+1}\!\!\iff\! 5 < 5^{n+1}\!-\!5^n = 4\cdot 5^n =: f(n).\,$ $\rm\:f(n)\:$ is increasing $\rm\: f(n\!+\!1)\ge f(n)\:$ hence a simple induction shows $\rm\:f(n) \ge f(1) > 5.$

Remark $\ $ This is a prototypical example of inequality telescoping. The hinted inductive proof - that an increasing function stays greater than its initial value - can be vividly viewed as collapsing $\rm\:f(n)\ge f(n\!-\!1)\ge\,\cdots\,f(2)\ge f(1)\:$ down to $\rm\:f(n)\ge f(1),\:$ using the transitivity of $\,\ge,\,$ just like one collapses the sections of a telescope down to one section.

Note that proving said lemma on increasing functions is actually easier then proving the simple case in your problem, because the special-case details no longer obscure the innate telescopic structure. Further, this lemma can be reused for many induction problems of similar type. Ditto for the higher-level conceptual understanding of how these types of induction work.

The above method of transforming an inductive proof into a much simpler telescopic proof is quite powerful, and very widely applicable, not only for inequalities but also for sums, products, etc. For many further examples see my prior posts on telescopy.

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Inequality holds for $n=1$.
For $n=k$ $$5^k + 5 < 5^{k+1}$$ We can write this as
$$5^{k+1} - (5^k + 5) = p$$ where $p$ is a positive integer.
$$5*5^k - 5^k - 5 = p$$ $$4*5^k - 5 = p$$ $$5^k=\frac{p+5}{4}$$ Now we have to prove it is true for $n = k+1$. So if $(5^{k+2} - (5^{k+1}+5))$ is positive then inequality holds true for $n=k+1$. $$5^{k+2} - (5^{k+1}+5)$$ $$25*5^k - (5*5^{k}+5)$$ $$25*5^k - 5*5^{k}-5$$ $$20*5^k - 5$$ substitute value of $5^k$, $$20*(\frac{p+5}{4}) - 5$$ $$5*(p+5)-5$$ $$5p+25-5$$ $$5p+20$$ $$5(p+4)$$ As $p$ is a positive integer, so $5(p+4)$ is also positive integer. Thus, we can say that $$5^{k+1}+5 < 5^{k+2}$$

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