\(u_{15}=16\cdot u_7\)
=>\(u_{10}\cdot q^5=16\cdot u_{10}\cdot q^{-3}\)
=>\(64\cdot q^5=16\cdot64\cdot q^{-3}\)
=>\(q^5=16\cdot q^{-3}\)
=>\(\dfrac{q^5}{q^{-3}}=16\)
=>\(q^8=16=\left(\sqrt{2}\right)^8\)
=>\(\left[{}\begin{matrix}q=\sqrt{2}\\q=-\sqrt{2}\end{matrix}\right.\)
TH1: \(q=\sqrt{2}\)
\(u_{10}=u_1\cdot q^9\)
=>\(u_1=\dfrac{64}{q^9}=\dfrac{64}{\left(\sqrt{2}\right)^9}=\sqrt{2}\)
\(s_{10}=\dfrac{u_1\cdot\left(1-q^{10}\right)}{1-q}=\dfrac{\sqrt{2}\cdot\left(1-\left(\sqrt{2}\right)^{10}\right)}{1-\sqrt{2}}=\dfrac{\sqrt{2}\cdot\left(1-32\right)}{1-\sqrt{2}}=-\sqrt{2}\left(\sqrt{2}+1\right)\left(-31\right)=31\sqrt{2}\left(\sqrt{2}+1\right)\)
\(s_{20}=\dfrac{u_1\cdot\left(1-q^{20}\right)}{1-q}=\dfrac{\sqrt{2}\cdot\left[1-\left(\sqrt{2}\right)^{20}\right]}{1-\sqrt{2}}\)
\(=\dfrac{-\sqrt{2}\cdot\left(1-2^{10}\right)}{\sqrt{2}-1}=-\sqrt{2}\left(\sqrt{2}+1\right)\left(1-1024\right)=1023\sqrt{2}\left(\sqrt{2}+1\right)\)
TH2: \(q=-\sqrt{2}\)
\(u_1=\dfrac{u_{10}}{q^9}=\dfrac{64}{\left(-\sqrt{2}\right)^9}=-\sqrt{2}\)
\(s_{10}=\dfrac{u_1\cdot\left(1-q^{10}\right)}{1-q}=\dfrac{-\sqrt{2}\left[1-\left(-\sqrt{2}\right)^{10}\right]}{1-\left(-\sqrt{2}\right)}\)
\(=\dfrac{-\sqrt{2}\left(1-2^5\right)}{1+\sqrt{2}}=-\sqrt{2}\left(\sqrt{2}-1\right)\left(1-2^5\right)=31\sqrt{2}\left(\sqrt{2}-1\right)\)
\(s_{20}=\dfrac{u_1\cdot\left(1-q^{20}\right)}{1-q}=\dfrac{-\sqrt{2}\cdot\left[1-\left(-\sqrt{2}\right)^{20}\right]}{1-\left(-\sqrt{2}\right)}\)
\(=\dfrac{-\sqrt{2}\left(1-2^{10}\right)}{1+\sqrt{2}}=\dfrac{\sqrt{2}\cdot\left(1024-1\right)}{\sqrt{2}+1}=1023\sqrt{2}\left(\sqrt{2}-1\right)\)
