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Question Number 121694 by bemath last updated on 11/Nov/20

$$Evaluate\underset{k=1}{\sum ^n}\frac{4k+\sqrt{4k^2-1}}{\sqrt{2k-1}+\sqrt{2k+1}}?$$

Answered by liberty last updated on 11/Nov/20

$$\mathrm{let}\sqrt{2\mathrm{k}-1}=\mathrm{a}\mathrm{and}\sqrt{2\mathrm{k}+1}=\mathrm{b}$$$$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=4\mathrm{k}\wedge \mathrm{ab}=\sqrt{{4\mathrm{k}}^2-1}$$$$\Rightarrow \frac{4\mathrm{k}+\sqrt{{4\mathrm{k}}^2-1}}{\sqrt{2\mathrm{k}-1}+\sqrt{2\mathrm{k}+1}}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+\mathrm{ab}}{\mathrm{a}+\mathrm{b}}$$$$\Rightarrow \frac{\frac{{\mathrm{a}}^3-{\mathrm{b}}^3}{\mathrm{a}-\mathrm{b}}}{\mathrm{a}+\mathrm{b}}=\frac{{\mathrm{a}}^3-{\mathrm{b}}^3}{{\mathrm{a}}^2-{\mathrm{b}}^2}=\frac{{\mathrm{a}}^3-{\mathrm{b}}^3}{-2}=\frac{{\mathrm{b}}^3-{\mathrm{a}}^3}{2}.\mathrm{since}$$$${\mathrm{a}}^2-{\mathrm{b}}^2=(2\mathrm{k}-1)-(2\mathrm{k}+1)=-2$$$$\Rightarrow \frac{4\mathrm{k}+\sqrt{{4\mathrm{k}}^2-1}}{\sqrt{2\mathrm{k}-1}+\sqrt{2\mathrm{k}+1}}=\frac{\sqrt{{(2\mathrm{k}+1)}^3}-\sqrt{{(2\mathrm{k}-1)}^3}}{2}$$$$\mathrm{then}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{4\mathrm{k}+\sqrt{{4\mathrm{k}}^2-1}}{\sqrt{2\mathrm{k}-1}+\sqrt{2\mathrm{k}+1}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{\sqrt{{(2\mathrm{k}+1)}^3}-\sqrt{{(2\mathrm{k}-1)}^3}}{2}$$$$=\frac{\sqrt{{(2\mathrm{n}+1)}^3}-1}{2}.\left[\mathrm{telescoping}\mathrm{series}\right]$$

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