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$$\frac{1}{9\sqrt11 + 11\sqrt9} + \frac{1}{11\sqrt13 + 13\sqrt11} + \frac{1}{13\sqrt15 + 15\sqrt13} + \ldots + \frac{1}{n\sqrt{n+2} + (n+2)\sqrt{n}} = \frac{1}{9}$$

Find the value of $n$.

I got the summation as $$\sum_{r=1}^n{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} = \sum_{r=1}^n{\frac{1}{\sqrt{2r+7}\sqrt{2r+9}(\sqrt{2r+7} + \sqrt{2r+9})}}$$ but I'm not able to simplify it further. I think I should've expressed it as difference of two summations but I'm not able to figure out which summations. Please help.

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  • 2
    $\begingroup$ Ok, Welcome to MathSE. Can you show your attempts for to solve this problem? $\endgroup$
    – user798113
    Commented Nov 11, 2020 at 15:13
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    $\begingroup$ Hint: the sum telescopes $\endgroup$
    – achille hui
    Commented Nov 11, 2020 at 15:18
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    $\begingroup$ Be careful about the use of $n$ here; as I read the opening formula, $n$ is not the number of terms being summed, but the (odd) number that appears in the final square root. In your summation version, that final square root is $\sqrt{2n+7}$, not $\sqrt n$. $\endgroup$
    – Barry Cipra
    Commented Nov 11, 2020 at 15:39
  • $\begingroup$ @BarryCipra so would the summation be from r = 1 to (n-7)/2? Or would the summation be represented in a different way altogether? $\endgroup$
    – user1729
    Commented Nov 11, 2020 at 15:56
  • $\begingroup$ @RanaMaihar, it might be better to write the final denominator in the opening expression as $(2n+7)\sqrt{2n+9}+(2n+9)\sqrt{2n+7}$ instead of $n\sqrt{n+2}+(n+2)\sqrt n$. That would make it consistent with using $n$ as the number of terms in the sum. $\endgroup$
    – Barry Cipra
    Commented Nov 11, 2020 at 16:02

2 Answers 2

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$$S_m=\sum_{r=1}^m{\frac{1}{(2r+7)\sqrt{2r+9} + (2r+9)\sqrt{2r+7}}} =\sum_{r=1}^m{\frac{(2r+7)\sqrt{2r+9} - (2r+9)\sqrt{2r+7}}{(2r+7)^2(2r+9)-(2r+9)^2(2r+7)}}$$ $$= \sum_{r=1}^m{\frac{(2r+7)\sqrt{2r+9} - (2r+9)\sqrt{2r+7}}{-2(2r+7)(2r+9)}} = -\frac{1}{2}\sum_{r=1}^m{\frac{\sqrt{2r+9}}{2r+9}} - {\frac{\sqrt{2r+7}}{2r+7}}$$ $$=\frac{1}{2}\sum_{r=1}^m{\frac{1}{\sqrt{2r+7}}} - {\frac{1}{\sqrt{2r+9}}} = \frac{1}{2}\left({\frac{1}{3}} - {\frac{1}{\sqrt{2m+9}}}\right) $$

So $$S_m = \frac{1}{9} \Longleftrightarrow {\frac{1}{\sqrt{2m+9}}} = \frac{1}{9} \Longleftrightarrow \sqrt{2m+9} = 9 \Longleftrightarrow m=36$$

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  • $\begingroup$ @cosmo5 My answer gives the value of $n$ such that the sum $S_n$ defined with the symbol $\sum$ is equal to $1/9$. This version is better because is it valid for all $n$, and you must not assume that $n$ is odd as in the sum with $+...+$. Of course it is easy to go from one to the other, so the OP will adapt this solution if he wants. $\endgroup$
    – TheSilverDoe
    Commented Nov 11, 2020 at 15:53
  • $\begingroup$ @cosmo5 I am not sure that there is a possible confusion ! I denoted $S_n$ the sum with the symbol $\sum$, and I worked with it... Everything is consistent (unless I have been mistaken, but I don't think so). I am just saying that this sum $S_n$ is equal to $1/9$ iff $n=36$. I am not talking about the sum with $+...+$. $\endgroup$
    – TheSilverDoe
    Commented Nov 11, 2020 at 15:56
  • $\begingroup$ $n$ is given in the original problem so if you had instead used $m$ then there wouldn't be any confusion at all. $\endgroup$
    – Neat Math
    Commented Nov 11, 2020 at 16:04
  • $\begingroup$ @TheSilverDoe, as I have pointed out in comments beneath the OP, there is confusion regarding the meaning of the variable $n$. Note that your answer and cosmo5's agree in the sense that $79=2\cdot36+7$. $\endgroup$
    – Barry Cipra
    Commented Nov 11, 2020 at 16:05
  • $\begingroup$ @NeatMath Fair point, I edited. $\endgroup$
    – TheSilverDoe
    Commented Nov 11, 2020 at 16:06
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Observe $$\dfrac{1}{9\sqrt{11} + 11\sqrt9}$$

$$ =\dfrac{1}{\sqrt 9 \sqrt {11} }\cdot\dfrac{1}{\sqrt9 + \sqrt{11}}$$

$$ =\dfrac{1}{\sqrt 9 \sqrt {11} }\cdot \dfrac{\sqrt{11} - \sqrt{9}}{2}$$

$$ =\dfrac{1}{2}\cdot \dfrac{\sqrt{11} - \sqrt{9}}{\sqrt 9 \sqrt {11} }$$

$$ =\dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{11}}\Big)$$

Thus LHS

$$ \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{11}}\Big) + \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{13}} - \dfrac{1}{\sqrt{11}}\Big) + \ldots + \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+2}}\Big)$$

telescopes to

$$ \dfrac{1}{2}\cdot \Big(\dfrac{1}{\sqrt{9}} - \dfrac{1}{\sqrt{n+2}}\Big) = \dfrac{1}{9}$$

$$ \Rightarrow \dfrac{1}{3} - \dfrac{1}{\sqrt{n+2}} = \dfrac{2}{9}$$

$$ \Rightarrow n+2 = 81 $$

$$ \Rightarrow \boxed{n=79} $$

Note The $n$ in your question and the $n$ in your attempt are different. I have found $n$ in your question. Indeed final term is $\frac{1}{\sqrt{81}}$ as in the other answer. The $n$ in your attempt is $36$. Indeed $2\cdot36+9=81$.

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