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I have the following problem:

Homework Problem

So I start as follows:

$B_{2}M_{2}=\frac{1}{\sqrt{3}}$ and I realize that $B_{2}M_{2} = A_{2}B_{2}$, so $B_{3}M_{3} = \frac{1}{\sqrt{3}}^{2}$. Next, I compute $A_{1}B_{2}=A_{1}B_{1}-B_{2}M_{2} = 1-\frac{1}{\sqrt{3}}$. For $A_{2}B_{3}$, we can do something similar, namely: $A_{2}B_{3} = A_{2}B_{2}-B_{3}M_{3}=\frac{1}{\sqrt{3}}-(\frac{1}{\sqrt{3}})^{2}$.

We can now derive the formula for computing the area of both triangles: $$\frac{1}{2}\sqrt{(A_{1}B_{1})^{2}+(B_{1}M_{1})^{2}}\cdot A_{1}B_{2} \sin(30)$$

Now since $A_{1}B_{1}=B_{1}M_{1}$, we can simplify it to: $$\frac{1}{\sqrt{2}}\sqrt{(A_{1}B_{1})^{2}}\cdot A_{1}B_{2}$$

This can be written as: $$S_{n} = \frac{1}{\sqrt{2}}\sqrt{(A_{n}B_{n})^{2}}\cdot A_{n}B_{n+1}$$

I fill in the missing pieces and I get: $$S_{n} = \frac{1}{\sqrt{2}} \cdot \sqrt{\left(\frac{1}{3}\right)^{n-1}} \cdot \left( \left(\frac{1}{3} \right) ^{n-1} - \left( \frac{1}{3} \right)^{n} \right) $$

Now I simplify it to:

$$S_n = \sqrt{6} \cdot 3^{-1.5n}$$

Now I have a small problem, if I try to solve $\sum_{n=1}^{\infty} \sqrt{6} \cdot 3^{-1.5n}$, I know we can rewrite it to $\sqrt{6} \sum_{n=1}^{\infty} 3^{-1.5n}$, but I have no idea how to continue to get the closed form.

In addition, I was wondering, is there a faster way to get the solution?

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  • $\begingroup$ Hint: $(3^{-1.5})^n=a^n$ and use geometric series. $\endgroup$ – MrYouMath Jul 13 '16 at 21:58
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where does this come from?

$\frac{1}{2}\sqrt{(A_{1}B_{1})^{2}+(B_{1}M_{1})^{2}}\cdot A_{1}B_{2} \sin(30)$

I think you want $\sin 45$. But that is overly complicated. $\frac 12 bh$ will suffice with $b = A_1B_2$ and $h = A_1B_1$ and there are 2 of these triangles, so you can double that.

$S_1 = 1 - \frac 1{\sqrt{3}}\\ S_n = (\frac 1{\sqrt 3}) S_{n-1}\\ S_n = (\frac 1{\sqrt 3})^{n-1} S_{1}\\ \sum_\limits{n=1}^{\infty} S_n = S_1\sum_\limits{n=0}^{\infty} (\frac 1{\sqrt 3})^n = \frac {S_1}{(1-\frac 1{\sqrt 3})} = 1$

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  • $\begingroup$ Yeah, I tried to use the triangle $A_{1}M_{1}B_{2}$, but I use $A_{1}M_{1}$ as base, so your solution is much more elegant! I also made a mistake and I was planning to use $\sin(45)$. PS: I realize that you asked where does this come from, but you can also mean the problem itself. It comes from a Korean College Exam (see: askakorean.blogspot.nl/2011/11/…). $\endgroup$ – Snowflake Jul 13 '16 at 22:24
  • $\begingroup$ But I have another problem... apparently the solution is not in my multiple choice right? $\endgroup$ – Snowflake Jul 13 '16 at 22:26
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Hint. One may recall the evaluation of a geometric series $$ \sum_{n=1}^\infty r^n=\frac{r}{1-r}, \quad |r|<1. $$ Then put $r=3^{-1.5}=0.1924\cdots$ to get the answer.

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  • $\begingroup$ Oh of course, thanks Olivier! Kind of ashamed that I forgot that one! $\endgroup$ – Snowflake Jul 13 '16 at 22:19

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