Results 1 to 5 of 5

Math Help - Infinite Sum, involving sqrt and a+b+c

  1. #1
    Member
    Joined
    Mar 2011
    Posts
    99

    Thumbs up Infinite Sum, involving sqrt and a+b+c

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,824
    Thanks
    713
    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

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2011
    Posts
    99
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: July 7th 2011, 12:09 AM
  2. 1/sqrt(x) - finite area, infinite volume?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 6th 2010, 07:02 PM
  3. proof of infinite n's s.t. sqrt(n) = irrational
    Posted in the Number Theory Forum
    Replies: 8
    Last Post: October 8th 2009, 01:46 PM
  4. Replies: 3
    Last Post: May 5th 2009, 02:38 PM
  5. Replies: 11
    Last Post: January 6th 2008, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum