Chương 3: DÃY SỐ. CẤP SỐ CỘNG VÀ CẤP SỐ NHÂN

\(u_{n+1}-u_n\)

\(=\sqrt{u_n+2}-u_n\)

\(=\dfrac{u_n+2-u_n^2}{\sqrt{u_n+2}+u_n}=\dfrac{-\left(u_n-2\right)\left(u_n+1\right)}{\sqrt{u_n+2}+u_n}\)

\(u_{n+1}=\sqrt{u_n+2}\)

=>\(u_{n+1}^2=u_n+2\)

=>\(u_{n+1}^2-4=u_n-2\)

=>\(\left(u_{n+1}-2\right)\left(u_{n+1}+2\right)=u_n-2\)

Để \(u_n< 2\) thì \(u_n-2< 0\)

=>\(u_{n+1}-2< 0\)

=>\(u_n< 2\forall n>=1\)

=>\(u_{n+1}-u_n>0\)

=>Đây là dãy tăng và bị chặn trên bởi 2

\(u_n=\dfrac{n^2+1}{2n^2-3}\)

\(=\dfrac{1}{2}\cdot\dfrac{n^2+1}{n^2-1,5}\)

\(=\dfrac{1}{2}\left(\dfrac{n^2-1,5+2,5}{n^2-1,5}\right)=\dfrac{1}{2}\left(1+\dfrac{2.5}{n^2-1,5}\right)< \dfrac{1}{2}\)

=>(U_n) là dãy số bị chặn

\(u_2=u_1^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^{2^{2-1}}\)

\(u_3=u_2^2=\left[\left(\dfrac{1}{2}\right)^2\right]^2=\left(\dfrac{1}{2}\right)^4=\left(\dfrac{1}{2}\right)^{2^{3-1}}\)

\(u_4=u_3^2=\left[\left(\dfrac{1}{2}\right)^4\right]^2=\left(\dfrac{1}{2}\right)^8=\left(\dfrac{1}{2}\right)^{2^{4-1}}\)

...

=>\(u_n=\left(\dfrac{1}{2}\right)^{2^{n-1}}\)

a: \(u_1=1;u_2=2;u_3=3;u_4=4;u_5=5;u_6=0;u_7=1\)

b: m=6k+a;n=6c+d

\(u_m=u_n\)

=>a=d

=>\(m=6k+a;n=6c+a\)

\(\left|m-n\right|=\left|6k+a-6c-a\right|=\left|6k-6c\right|=6\left|k-c\right|⋮6\)

Để \(cos\left(x-\dfrac{\Omega}{4}\right);sinx;cos\left(x+\dfrac{\Omega}{4}\right)\) là ba số hạng liên tiếp của cấp số nhân thì \(sin^2x=cos\left(x-\dfrac{\Omega}{4}\right)\cdot cos\left(x+\dfrac{\Omega}{4}\right)\)

=>\(sin^2x=\sqrt{2}\left(cosx-sinx\right)\cdot\sqrt{2}\left(cosx+sinx\right)\)

=>\(sin^2x=2cos^2x-2sin^2x\)

=>\(3\cdot sin^2x=2\cdot cos^2x\)

=>\(\dfrac{sin^2x}{cos^2x}=\dfrac{2}{3}\)

=>\(tan^2x=\dfrac{2}{3}\)

=>\(\left[{}\begin{matrix}tanx=\dfrac{\sqrt{6}}{3}\\tanx=-\dfrac{\sqrt{6}}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(\dfrac{\sqrt{6}}{3}\right)+k\Omega\\x=arctan\left(-\dfrac{\sqrt{6}}{3}\right)+k\Omega\end{matrix}\right.\)

\(q=\dfrac{u_{n+1}}{u_n}=\dfrac{9}{81}=\dfrac{1}{9}\)