Root series

Algebra Level 3

Determine the value of

$\small \frac {\sqrt {3^2-4}+1^2-2^2} {\sqrt {4^2-1}}+\frac {\sqrt {5^2-4}+2^2-3^2} {\sqrt {6^2-1}}+\frac {\sqrt {7^2-4}+3^2-4^2} {\sqrt {8^2-1}}+\cdots+\frac {\sqrt {79^2-4}+39^2-40^2} {\sqrt {80^2-1}}\\$

\frac {3\sqrt 3-\sqrt {79}} 3

\frac {3\sqrt 3+\sqrt {79}} 3

\frac {3\sqrt 3+\sqrt {79}} 9

\frac {3\sqrt 3-\sqrt {79}} 9

1 solution

Chew-Seong Cheong
Feb 13, 2021

The given sum can be written as:

$\begin{aligned} S & = \sum_{n=1}^{39} \frac {\sqrt{(2n+1)^2-4}+n^2-(n+1)^2}{\sqrt{(2(n+1))^2 - 1}} & \small \blue{\text{Note that }a^2 - b^2 = (a+b)(a-b)} \\ & = \sum_{n=1}^{39} \frac {\sqrt{(2n+1+2)(2n+1-2)}-(2n+1)}{\sqrt{(2n+2+1)(2n+2-1)}} \\ & = \sum_{n=1}^{39} \frac {\sqrt{(2n+3)(2n-1)}-(2n+1)}{\sqrt{(2n+3)(2n+1)}} \\ & = \sum_{n=1}^{39} \left(\sqrt{\frac {2n-1}{2n+1}} - \sqrt{\frac {2n+1}{2n+3}} \right) \\ & = \sqrt{\frac 13} - \cancel{\sqrt{\frac 35}} + \cancel{\sqrt{\frac 35}} - \cancel{\sqrt{\frac 57}} + \cancel{ \sqrt{\frac 57}} - \cancel{\sqrt{\frac 79}} + \cdots + \cancel{\sqrt{\frac {77}{79}}} - \sqrt{\frac {79}{81}} \\ & = \sqrt{\frac 13} - \sqrt{\frac {79}{81}} \\ & = \boxed{\frac {3\sqrt3 - \sqrt{79}}9} \end{aligned}$

Root series

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