Cho dãy số (un) với un = (6.5^n + 5.2^n)/(5^n + 2^n). Khi đó tổng S = 1/(u1−5) + 1/(u2−5)

Ta có  ${\mathrm{u}}_{\mathrm{n}}-5=\frac{6{.5}^{\mathrm{n}}+5{.2}^{\mathrm{n}}}{5^{\mathrm{n}}+2^{\mathrm{n}}}-5=\frac{5^{\mathrm{n}}}{5^{\mathrm{n}}+2^{\mathrm{n}}}$$\Rightarrow \frac{1}{{\mathrm{u}}_{\mathrm{n}}-5}=\frac{5^{\mathrm{n}}+2^{\mathrm{n}}}{5^{\mathrm{n}}}=1+{\left(\frac{2}{5}\right)}^{\mathrm{n}}$

$$\begin{gathered} \Rightarrow \mathrm{S}=\frac{1}{{\mathrm{u}}_1-5}+\frac{1}{{\mathrm{u}}_2-5}+...+\frac{1}{{\mathrm{u}}_{2021}-5} \\ =\left[1+{\left(\frac{2}{5}\right)}^1\right]+\left[1+{\left(\frac{2}{5}\right)}^2\right]+...+ \\ \left[1+{\left(\frac{2}{5}\right)}^{2021}\right] \\ =\left[2021+\frac{2}{5}.\frac{1-{\left(\frac{2}{5}\right)}^{2021}}{1-\frac{2}{5}}\right] \\ =\left[2021+\frac{2}{3}.\left(1-{\left(\frac{2}{5}\right)}^{2021}\right)\right] \\ =\left[\frac{6065}{3}-\frac{2}{3}{\left(\frac{2}{5}\right)}^{2021}\right]. \end{gathered}$$

Mà $\mathrm{S}=\frac{\mathrm{a}}{3}-\frac{\mathrm{b}}{3}{\left(\frac{2}{5}\right)}^{\mathrm{c}}$  nên $\mathrm{a}=6065;$$\mathrm{b}=2;$$\mathrm{c}=2021$.

Nên $\mathrm{a}+2{\mathrm{b}}^2-2\mathrm{c}=6065+2{.2}^2-2.2021$$=2031$