Bài 2: Dãy số

r có đáp án chx

Do \(n\in N\Rightarrow n+1>0\Rightarrow u_n=\dfrac{1}{n+1}>0\Rightarrow u_n\) bị chặn dưới bởi 0

Do \(n\in N\Rightarrow n\ge0\Rightarrow n+1\ge1\Rightarrow\dfrac{1}{n+1}\le1\)

\(\Rightarrow u_n\) bị chặn trên bởi 1

\(\Rightarrow\) Dãy số đã cho là dãy bị chặn

\(u_n-4u_{n-1}+3u_{n-2}=5.2^n\)

\(\Leftrightarrow u_n-u_{n-1}-3\left(u_{n-1}-u_{n-2}\right)=5.2^n\)

Đặt \(u_n-u_{n-1}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1-u_0=4\\v_n-3v_{n-1}=5.2^n\end{matrix}\right.\)

\(\Rightarrow v_n+10.2^n=3\left(v_{n-1}+10.2^{n-1}\right)\)

Đặt \(v_n+10.2^n=x_n\Rightarrow\left\{{}\begin{matrix}x_1=v_1+10.2^1=24\\x_n=3x_{n-1}\end{matrix}\right.\)

\(\Rightarrow x_n\) là CSN với công bội 3

\(\Rightarrow x_n=24.3^{n-1}\)

\(\Rightarrow v_n=x_n-10.2^n=24.3^{n-1}-10.2^n=8.3^n-10.2^n\)

\(\Rightarrow u_n-u_{n-1}=8.3^n-10.2^n\)

\(\Rightarrow u_n-12.3^n+20.2^n=u_{n-1}-12.3^{n-1}+20.2^{n-1}\)

Đặt \(u_n-12.3^n+20.2^n=y_n\Rightarrow\left\{{}\begin{matrix}y_1=u_1-12.3^1+20.2^1=7\\y_n=y_{n-1}=...=y_1=7\end{matrix}\right.\)

\(\Rightarrow u_n=12.3^n-20.2^n+7\)

Với \(n=0\Rightarrow0-0+0-0+0-0=0⋮24\left(đúng\right)\)

Với \(n=1\Rightarrow1-3+6-7+5-2=0⋮24\left(đúng\right)\)

G/s \(n=k\Rightarrow\left(k^6-3k^5+6k^4-7k^3+5k^2-2k\right)⋮24\)

\(\Rightarrow k\left(k^5-3k^4+6k^3-7k^2+5k-2\right)⋮24\\ \Rightarrow k\left(k+1\right)\left(k^2+k+1\right)\left(k^2-k+2\right)⋮24\)

Với \(n=k+1\), ta cần cm \(\left[\left(k+1\right)^6-3\left(k+1\right)^5+6\left(k+1\right)^4-7\left(k+1\right)^3+5\left(k+1\right)^2-2\left(k+1\right)\right]⋮24\)

Ta có \(\left(k+1\right)^6-3\left(k+1\right)^5+6\left(k+1\right)^4-7\left(k+1\right)^3+5\left(k+1\right)^2-2\left(k+1\right)\)

\(=\left(k+1\right)\left[\left(k+1\right)^5-3\left(k+1\right)^4+6\left(k+1\right)^3-7\left(k+1\right)+5\left(k+1\right)-2\right]\\ =\left(k+1\right)\left(k+1-1\right)\left[\left(k+1\right)^2-\left(k+1\right)+1\right]\left[\left(k+1\right)^2-\left(k+1\right)+2\right]\\ =k\left(k+1\right)\left(k^2+k+1\right)\left(k^2+k+2\right)\)

Mà theo GT quy nạp ta có \(k\left(k+1\right)\left(k^2+k+1\right)\left(k^2+k+2\right)⋮24\)

Vậy ta được đpcm

\(u_1=1\)

\(u_2=1\)

\(u_3=u_2+u_1=1+1=2\)

\(u_4=u_3+u_2=2+1=3\)

\(u_5=u_4+u_3=3+2=5\)