Infinite Sum, involving sqrt and a+b+c

**Zellator** · June 8th 2011, 03:01 PM

Hi Forum!

I've found this strange question, which asks the following:

If $\sqrt{10+\sqrt{10+...}}= (a+\sqrt{b})/c$
Find a+b+c

Now, this question was in a quadratics exam.
The solution given is quite simple, but it doesn't really seem to make sense.

Can someone help?
Thanks

**TheCoffeeMachine** · June 8th 2011, 03:28 PM

I'm not sure how they expect you to find a+b+c, but here's a start:

$\begin{aligned} & \left(\frac{a+\sqrt{b}}{c}\right) = \sqrt{10+\sqrt{10+\sqrt{10+\cdots}}} = \sqrt{10+\left(\frac{a+\sqrt{b}}{c}\right) \right) \\ & \therefore \left(\frac{a+\sqrt{b}}{c}\right)^2 = 10+\left(\frac{a+\sqrt{b}}{c}\right) \Rightarrow \left(\frac{a+\sqrt{b}}{c}\right)^2-\left(\frac{a+\sqrt{b}}{c}\right)-10 = 0. \end{aligned}$

$t^2-t-10 = \left(t-\frac{1}{2}\right)^2-\frac{41}{4} = \left(t-\frac{1}{2}-\frac{\sqrt{41}}{2}\right)\left(t-\frac{1}{2}+\frac{\sqrt{41}}{2}\right)$ .

$\therefore ~ \frac{a+\sqrt{b}}{c} =\frac{1}{2}\left(1 \pm \sqrt{41}\right) \Rightarrow \frac{a+\sqrt{b}}{c} = \frac{1}{2}\left(1 + \sqrt{41}\right)$ .

**Soroban** · June 8th 2011, 09:32 PM

Hello, Zellator!

Another approach . . .

$\text{Evaluate: }\:\sqrt{10+\sqrt{10+\sqrt{10 + \hdots}}}$

$\text{Write the answer in the form: }\:\frac{a+\sqrt{b}}{c} \;\;\;\text{and find }a+b+c.$

$\text{Let: }\:x \;=\;\sqrt{10 + \underbrace{\sqrt{10 + \sqrt{10 + \hdots}}}_{\text{This is }x}}$

$\text{We have: }\:x \;=\;\sqrt{10 + x} \quad\Rightarrow\quad x^2 \;=\;10 + x \quad\Rightarrow\quad x^2 - x - 10 \:=\:0$

$\text{Quadratic Formula: }\:x \;=\;\frac{1 \pm\sqrt{41}}{2}$

$\text{Since }x\text{ is positive: }\:x \;=\;\frac{1 + \sqrt{41}}{2}$
$\text{We have: }\:\begin{Bmatrix}a &=& 1 \\ b &=& 41 \\ c &=& 2\end{Bmatrix}$

$\text{Therefore: }\;a + b + c \;=\; 1 + 41 + 2 \;=\;44$

**chisigma** · June 8th 2011, 10:12 PM

Is has been demonstrated that...

$\sqrt{10+\sqrt{10 + ...}}= \frac{1+\sqrt{41}}{2}$ (1)

... so that is...

$\frac{a + \sqrt{b}}{c} = \frac{1+\sqrt{41}}{2}$ (2)

Now, supposing that a, b and c are integers, the (2) has infinite solutions... the most obvious is...

$a=1, b=41, c=2 \implies a + b + c = 44$

Also solution however is...

$a=2, b= 164, c=4 \implies a + b + c = 170$

... and we can proceed indefinitely

...

Kind regards

$\chi$ $\sigma$

**Zellator** · June 9th 2011, 01:17 PM

Thanks all for your replies!!!
It's great to see people willing to help!
chisigma, the good thing is that this is a past test question with multiple choices.
So, if 44 wasn't there we could also say that 170 would be a result!

Thanks Soroban, with your clear-as-always explanations.

And CoffeeMachine for the introduction to it.
Before asking I was stuck on the second line.

Thanks all for your cooperation

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