Cho dãy số (un), biết un = 2^(1/(n^2 + 3n + 2)), mọi n thuộc N sao. Tích của 2021 số hạng đầu tiên bằng

Ta có:

$$\begin{gathered} \frac{1}{{\mathrm{n}}^2+3\mathrm{n}+2}=\frac{1}{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)} \\ =\frac{1}{1.2-1.1}\left(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+2}\right) \\ =\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+2} \end{gathered}$$

Suy ra: ${\mathrm{u}}_{\mathrm{n}}=2^{\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+2}}$ .

$\Rightarrow$ $\left\{\begin{array}{l}{\mathrm{u}}_1=2^{\frac{1}{2}-\frac{1}{3}}\\ {\mathrm{u}}_2=2^{\frac{1}{3}-\frac{1}{4}}\\ {\mathrm{u}}_3=2^{\frac{1}{4}-\frac{1}{5}}\\ .....\\ {\mathrm{u}}_{2021}=2^{\frac{1}{2022}-\frac{1}{2023}}\end{array}\right.$.

$$\begin{gathered} \Rightarrow {\mathrm{u}}_1.{\mathrm{u}}_2.{\mathrm{u}}_3....{\mathrm{u}}_{2021} \\ =2^{\frac{1}{2}-\frac{1}{3}}.2^{\frac{1}{3}-\frac{1}{4}}.2^{\frac{1}{4}-\frac{1}{5}}.{...2}^{\frac{1}{2022}-\frac{1}{2023}} \\ =2^{\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2022}-\frac{1}{2023}} \\ =2^{\frac{1}{2}-\frac{1}{2023}}=2^{\frac{2021}{4046}} \end{gathered}$$

Chọn đáp án C.