Cho cấp số nhân (un) thỏa mãn . Tính tổng s2011

* Hướng dẫn giải

Chọn C

Gọi q là công bội của cấp số. Khi đó ta có

$$\begin{gathered} \left\{\begin{array}{l}{u}_1+u_2+\hspace{0.17em}{u}_3+\hspace{0.17em}{u}_4+\hspace{0.17em}{u}_5= 11\\ u_1 +\hspace{0.17em}{u}_5= \frac{82}{11}\end{array}\right.\iff \left\{\begin{array}{l}{u}_2+\hspace{0.17em}{u}_3+\hspace{0.17em}{u}_4= \frac{39}{11}\\ u_1 +\hspace{0.17em}{u}_5= \frac{82}{11}\end{array}\right. \\ \iff \left\{\begin{array}{l}{u}_1(q +\hspace{0.17em}{q}^2+\hspace{0.17em}{q}^3) = \frac{39}{11} \left(1\right)\hspace{0.17em}\\ u_1( 1+\hspace{0.17em}{q}^4) = \frac{82}{11} \left(2\right)\end{array}\right. \end{gathered}$$

Lấy (2) chia (1) ta được:

$$\begin{gathered} \frac{1 +\hspace{0.17em}{q}^4}{q +\hspace{0.17em}{q}^2+\hspace{0.17em}{q}^3}= \frac{82}{39}\iff 39 +\hspace{0.17em}39q^4= 82q +\hspace{0.17em}82q^2+\hspace{0.17em}82q^3 \\ \iff 39q^4- 82q^3-\hspace{0.17em}82q^2+\hspace{0.17em}82q- 39= 0 \\ \iff q = 3; q = \frac{1}{3} \end{gathered}$$

Ta có $S_{2011}=u_1\frac{q^{2011}-1}{q-1}$

+ Với $q=\frac{1}{3}\Rightarrow u_1= \frac{81}{11}\Rightarrow S_{2011}=\frac{243}{22}\left(1-\frac{1}{3^{2011}}\right)$

+ Với $q=3\Rightarrow u_1= \frac{1}{11}\Rightarrow S_{2011}=\frac{1}{22}\left(3^{2011}-1\right)$