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Question Number 137588 by bramlexs22 last updated on 04/Apr/21

$$Forapositivenumbern,let$$$$f\left(n\right)bethevalueof$$$$f\left(n\right)=\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$$$$calculatef\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(40\right).$$

Answered by bemath last updated on 04/Apr/21

$$\iff f\left(n\right)=\frac{{\left(\sqrt{2n+1}\right)}^2+{\left(\sqrt{2n-1}\right)}^2+\sqrt{(2n+1)(2n-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}$$$$f\left(n\right)=\frac{{\left(\sqrt{2n+1}\right)}^3-{\left(\sqrt{2n-1}\right)}^3}{2}$$$$f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(40\right)$$$$=\frac{\sqrt{3^3}-\sqrt{1^3}}{2}+\frac{\sqrt{5^3}-\sqrt{3^3}}{2}+\frac{\sqrt{7^3}-\sqrt{5^3}}{2}+...+\frac{\sqrt{{81}^3}-\sqrt{{79}^3}}{2}$$$$\left[telescopyseries\right]$$$$f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(40\right)$$$$=\frac{\sqrt{{81}^3}-\sqrt{1^3}}{2}=\frac{9^3-1}{2}=\frac{728}{2}=364$$

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