Biết lim (1/6 + 1/12 + 1/24 +...+ 1/3.2n) = ab, với a, b thuộc N và a/b tối giản. Tính P = a − b^2

Ta có

$$\begin{gathered} \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{3{.2}^{\mathrm{n}}} \\ =\frac{1}{3.2}+\frac{1}{3{.2}^2}+\frac{1}{3{.2}^3}+...+\frac{1}{3{.2}^{\mathrm{n}}} \\ =\frac{1}{3}\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{\mathrm{n}}}\right) \\ =\frac{1}{3}.\frac{1}{2}.\frac{\frac{1}{2^{\mathrm{n}}}-1}{\frac{1}{2}-1}=\frac{1}{3}-\frac{1}{3}.\frac{1}{2^{\mathrm{n}}} \end{gathered}$$

Khi đó $\lim \left(\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{3{.2}^{\mathrm{n}}}\right)$$=\lim \left(\frac{1}{3}-\frac{1}{3}.\frac{1}{2^{\mathrm{n}}}\right)$$=\frac{1}{3}$  .

Ta có a = 1, b = 3. Vậy $\mathrm{P}=\mathrm{a}-{\mathrm{b}}^2=-8$.