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Question Number 144858 by mathdanisur last updated on 29/Jun/21

Answered by mindispower last updated on 29/Jun/21

$${(n^2+2n+1)}^{\frac{1}{3}}={(n+1)}^{\frac{2}{3}}=a^2$$$${(n^2-2n+1)}^{\frac{1}{3}}={\left({(n-1)}^2\right)}^{\frac{1}{3}}=b^2$$$${(n^2-1)}^{\frac{1}{3}}=\mathit{a}\mathit{b}$$$$\frac{1}{{\mathit{a}}^2+{\mathit{b}}^2+\mathit{a}\mathit{b}}=\frac{\mathit{a}-\mathit{b}}{{\mathit{a}}^3-{\mathit{b}}^3}=\frac{\sqrt[3]{\mathit{n}+1}-\sqrt[3]{\mathit{n}-1}}{2}$$$$\underset{n=1}{\sum ^{999}}\frac{\sqrt[3]{n+1}-\sqrt[3]{n-1}}{2}=\frac{\sqrt[3]{1000}+\sqrt[3]{999}-1}{2}=\frac{9+\sqrt[3]{999}}{2}$$$$$$

Commented by mathdanisur last updated on 29/Jun/21

$$alotperfectSirthankyou$$

Commented by mathdanisur last updated on 30/Jun/21

$$DearSir,butanswer5$$

Commented by mathdanisur last updated on 30/Jun/21

$$f\left(1\right)+f\left(2\right)+f\left(3\right)+...\mathit{n}\mathit{o},f\left(1\right)+f\left(3\right)+f\left(5\right)+...$$

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