Chain of square roots

Level 2

1 1 2 + 2 1 + 1 2 3 + 3 2 + 1 3 4 + 4 3 + ⋯ + 1 35 36 + 36 35 = ? \large \frac{1}{1\sqrt{2} + 2\sqrt{1}} + \frac{1}{2\sqrt{3} + 3\sqrt{2}} + \frac{1}{3\sqrt{4} + 4\sqrt{3}} + \dots + \frac{1}{35\sqrt{36} + 36\sqrt{35}} = ?

2 3 \frac{2}{3} 36 35 \frac{36}{35} 5 6 \frac{5}{6} 36 37 \frac{36}{37} 35 36 \frac{35}{36}

Frodo Baggins by Frodo Baggins , 21, New Zealand

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1 solution

S = 1 1 2 + 2 1 + 1 2 3 + 4 2 + 1 3 4 + 4 3 + ⋯ + 1 35 36 + 36 35 = ∑ n = 1 35 1 n n + 1 + ( n + 1 ) n = ∑ n = 1 35 ( n + 1 ) n − n n + 1 ( ( n + 1 ) n + n n + 1 ) ( ( n + 1 ) n − n n + 1 ) = ∑ n = 1 35 ( n + 1 ) n − n n + 1 ( n + 1 ) 2 n − n 2 ( n + 1 ) = ∑ n = 1 35 ( n + 1 ) n − n n + 1 n ( n + 1 ) = ∑ n = 1 35 ( 1 n − 1 n + 1 ) = ∑ n = 1 35 1 n − ∑ n = 2 36 1 n = 1 1 − 1 3 6 = 1 − 1 6 = 5 6 \begin{aligned} S & = \frac 1{1\sqrt 2 + 2\sqrt 1} + \frac 1{2\sqrt 3 + 4\sqrt 2} + \frac 1{3\sqrt 4 + 4\sqrt 3} + \cdots + \frac 1{35 \sqrt {36} + 36 \sqrt {35}} \\ & = \sum_{n=1}^{35} \frac 1{n\sqrt {n+1} + (n+1)\sqrt n} \\ & = \sum_{n=1}^{35} \frac {(n+1)\sqrt n - n\sqrt{n+1}}{\left((n+1)\sqrt n + n\sqrt{n+1}\right)\left((n+1)\sqrt n - n\sqrt{n+1}\right)} \\ & = \sum_{n=1}^{35} \frac {(n+1)\sqrt n - n\sqrt{n+1}}{(n+1)^2n - n^2(n+1)} \\ & = \sum_{n=1}^{35} \frac {(n+1)\sqrt n - n\sqrt{n+1}}{n(n+1)} \\ & = \sum_{n=1}^{35} \left(\frac 1{\sqrt n} - \frac 1{\sqrt{n+1}} \right) \\ & = \sum_{n=1}^{35} \frac 1{\sqrt n} - \sum_{n=2}^{36} \frac 1{\sqrt n} \\ & = \frac 1{\sqrt 1} - \frac 1{\sqrt 36} = 1 - \frac 16 = \boxed{\dfrac 56} \end{aligned}

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