Question

Tìm số các số hạng của cấp số nhân \(\left(u_n\right)\) biết :

a) \(q=2\)               \(u_n=96\)                \(S_n=189\)

b) \(u_1=2\)             \(u_n=\dfrac{1}{8}\)                 \(S_n=\dfrac{31}{8}\)

Answer 1

a) \(u_n=u_1.q^{n-1}=u_1.2^{n-1}\)
\(S_n=\dfrac{u_1\left(1-q^n\right)}{1-q}=\dfrac{u_1\left(1-2^n\right)}{1-2}=u_1\left(2^n-1\right)\);
\(\dfrac{S_n}{u_n}=\dfrac{u_1\left(2^n-1\right)}{u_1.2^{n-1}}=\dfrac{2^n-1}{2^{n-1}}=2-\dfrac{1}{2^{n-1}}=\dfrac{63}{32}\)
Vì vậy \(\dfrac{1}{2^{n-1}}=\dfrac{1}{32}\) \(\Leftrightarrow\dfrac{1}{2^{n-1}}=\dfrac{1}{2^5}\)\(\Leftrightarrow n-1=5\Leftrightarrow n=6\).
b)
\(u_n=2.q^{n-1}=\dfrac{1}{8}\)\(\Rightarrow q^{n-1}=\dfrac{1}{16}\)
\(S_n=\dfrac{2\left(1-q^n\right)}{1-q}=\dfrac{2\left(1-q.q^{n-1}\right)}{1-q}=\dfrac{2\left(1-\dfrac{1}{16}q\right)}{1-q}=\dfrac{31}{8}\);
Suy ra \(q=-1\).