# which one is bigger $100^n+99^n$ or $101^n$

Suppose $n \in \mathbb{N} , n>1000$ now how can we prove :which one is bigger $$100^n+99^n \text{ or } 101^n \text{ ? }$$ I tried to use $\log$ but get nothing . Then I tried for binomial expansion...but I get stuck on this .

can someone help me ? thanks in advance.

• $101^n$, by a mile! – Lord Shark the Unknown Aug 16 '17 at 20:37
• @LordShark "How can we say which is bigger"... I would hope that the OP already knows that $101^n$ is eventually larger, but it isn't immediately clear when that begins happening – JMoravitz Aug 16 '17 at 20:39
• I know $101^n$is bigger ,but i am looking for an idea to prove itt – Khosrotash Aug 16 '17 at 20:39
• It's not that easy... for $n\leq 48$ we have $99^n+100^n>101^n$ – Raffaele Aug 16 '17 at 21:54
• You should know that PSQ's are frowned upon and that you should include more context? This is why I've decided to downvote. If you showed, for example, some sign of attempt, I'll likely retract this vote. – Simply Beautiful Art Aug 29 '17 at 21:07

Of course $$1.01^n>1+\frac{n}{100}>2$$ for $n>100$, and obviously $1+0.99^n<2$.

• Can you do it for $n>50$, though? :-) – Joffan Aug 16 '17 at 20:46
• @Joffan Of course $\ln 2\approx 0.69$... – Lord Shark the Unknown Aug 16 '17 at 20:47
• Clever! I love it. – 伽罗瓦 Aug 16 '17 at 20:55

What happened when you tried binomial expansion?

$101^n = (100 + 1)^n = 100^n + n*100^{n-1} + ....$

So you said $n > 1000$ (!!!) so

$100^n + n*100^{n-1}> 100^n + 1000*100^{n-1} > 100^n + 100*100^{n-1} = 100^n + 100^n> 100^n + 99^n$.

By quite a bit! I'm not sure how you could have missed that if you tried the binomial expansion.

• While this answer is of course correct, I think the final line is uncalled for. – Noah Schweber Aug 16 '17 at 21:00
• Meh... maybe... – fleablood Aug 16 '17 at 21:01

$100^n + 99^n$ and $101^n$

Divide all terms by $101^n$

$\left(\dfrac{100}{101}\right)^n+\left(\dfrac{99}{101}\right)^n<2 \left(\dfrac{100}{101}\right)^n$

and $2 \left(\dfrac{100}{101}\right)^n<1$ when $\left(\dfrac{100}{101}\right)^n<\dfrac{1}{2}$

that is when $n\log\dfrac{100}{101}<\log\dfrac{1}{2}$

$n\geq 70$

so for $n>1000$ we have $100^n + 99^n < 101^n$

The binomial expansion of $(k+1)^n$ starts:

$$(k+1)^n = k^n+nk^{n-1}+\cdots$$

so $(k+1)^n > k^n+nk^{n-1}$ (with $k,n>1$).

With $n=99,$ we have:

\begin{align} 101^{99} &> 100^{99} + 99\cdot 100^{98}\\ &> 100^{99} + 99^{99} \\ \end{align}

Pushing this technique a little further, and still just using $(k+1)^n > k^n+nk^{n-1},$ we can take $n=62,$ to get:

\begin{align} 101^{62} &> 100^{62} + 62\cdot 100^{61}\\ &> 100^{62} + 62(99^{61} + 61\cdot99^{60})\\ &\quad= 100^{62} + 62\cdot 99^{61} + 3782\cdot 99^{60}\\ &> 100^{62} + 62\cdot 99^{61} + 37\cdot 99\cdot 99^{60}\\ &\quad= 100^{62} + (62+37)\cdot 99^{61}\\ \text{so}\quad101^{62} &> 100^{62} + 99^{62}\\ \end{align}

And of course $101^k>100^k+99^k \implies 101^{k+1} > 101(100^k+99^k) > 100^{k+1}+99^{k+1}$

Consider $$f(x)=100^x+99^x-101^x=101^x\left(\left(\frac{100}{101}\right)^x+\left(\frac{99}{101}\right)^x-1\right)$$ The last factor tends to $-1$ when $x\to\infty$ so $f(x)$ is negative from a certain value of $x$ and since $f(0)=1\gt 0$ and the functions are concave we have $$100^x+99^x\gt 101^x\text{ for } x\lt x_0\\100^x+99^x\lt 101^x\text{ for } x\gt x_0$$ where $x_0$ is such that $f(x_0)=0$.

Calculation gives $$48\lt x_0\lt 49$$

Thus $101^n$ is the greater for $n\gt 1000$

(Actually for $n\gt48$. Besides $100^n+99^n$ is greater for $0\le n\le 48$)

The left hand side of the following beomes monotonic decreasing after a certain n>m.

$(\frac{100}{101}) ^{n} + (\frac{99}{101} )^{n} <1 \;\; \forall n>m$