Cho dãy số

B

- Hướng dẫn giải

Ta có:

$$\begin{gathered} U_n=\frac{1}{\sqrt[4]{n^3}+\sqrt[4]{n^3+3n^2+3n+1}} \\ \frac{1}{\sqrt[4]{n^3+2n^2+n}+\sqrt[4]{n^3+n^2}} \\ =\frac{1}{\sqrt{n}\sqrt[4]{n}+\sqrt{n}\sqrt[4]{n+1}} \\ \frac{1}{\sqrt{n+1}\sqrt[4]{n}+\sqrt{n+1}\sqrt[4]{n+1}} \\ =\frac{1}{\sqrt{n}\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)} \\ \frac{1}{\sqrt{n+1}\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)} \\ =\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)} \\ \frac{1}{\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)} \\ =\frac{\sqrt[4]{n+1}-\sqrt[4]{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)} \\ \frac{1}{\left(\sqrt{n+1}-\sqrt{n}\right)} \\ =\sqrt[4]{n+1}-\sqrt[4]{n},n\ge 1 \end{gathered}$$

Khi đó

$$\begin{gathered} S=u_1+u_2+..+u_{{2018}^4-1} \\ =\sqrt[4]{2}-\sqrt[4]{1}+\sqrt[4]{3}-\sqrt[4]{2} \\ +..+\sqrt[4]{{2018}^4}-\sqrt[4]{{2018}^4-1} \\ =\sqrt[4]{{2018}^4}-1=2017 \end{gathered}$$

Đáp án B