Bài tập nâng cao · Bài 22
Bài tập nâng cao Dãy số · Bài 22
Cho dãy số $(u_n)$ định bởi: $u_n = \frac{a.n^{4}+2}{2n^{4}+5};n\in N*$. Định a để dãy số $(u_n)$tăng.
Xem lời giải
Lời giải
Ta có: $u_n = \frac{a.n^{4}+2}{2n^{4}+5} = \frac{a}{2}+\frac{4-5a}{2\left( 2n^{4}+5 \right)};n\in N*$
$u_{n+1}-u_n = \frac{4-5a}{2\left[ 2\left( n+1 \right)^{4}+5 \right]}-\frac{4-5a}{2\left[ 2n^{4}+5 \right]} = \frac{4-5a}{2}\left[ \frac{1}{2\left( n+1 \right)^{4}+5}-\frac{1}{2n^{4}+5} \right]$
$\Rightarrow u_{n+1}-u_n = \frac{4-5a}{2}\frac{2n^{4}+5-2\left( n+1 \right)^{4}-5}{\left[ 2\left( n+1 \right)^{4}+5 \right]\left[ 2\left( n+1 \right)^{4}+5 \right]}$
$= \left( 4-5a \right)\frac{n^{4}-\left( n+1 \right)^{4}}{\left[ 2\left( n+1 \right)^{4}+5 \right]\left[ 2\left( n+1 \right)^{4}+5 \right]}$
Mà: $\frac{n^{4}-\left( n+1 \right)^{4}}{\left[ 2\left( n+1 \right)^{4}+5 \right]\left[ 2\left( n+1 \right)^{4}+5 \right]}<0; \forall n\in N*$
Nên: $\left( u_n \right)$ tăng $\Leftrightarrow u_{n+1}-u_n>0; \forall n\in N*\Leftrightarrow 4-5a<0\Leftrightarrow a>\frac{4}{5}$
$u_{n+1}-u_n = \frac{4-5a}{2\left[ 2\left( n+1 \right)^{4}+5 \right]}-\frac{4-5a}{2\left[ 2n^{4}+5 \right]} = \frac{4-5a}{2}\left[ \frac{1}{2\left( n+1 \right)^{4}+5}-\frac{1}{2n^{4}+5} \right]$
$\Rightarrow u_{n+1}-u_n = \frac{4-5a}{2}\frac{2n^{4}+5-2\left( n+1 \right)^{4}-5}{\left[ 2\left( n+1 \right)^{4}+5 \right]\left[ 2\left( n+1 \right)^{4}+5 \right]}$
$= \left( 4-5a \right)\frac{n^{4}-\left( n+1 \right)^{4}}{\left[ 2\left( n+1 \right)^{4}+5 \right]\left[ 2\left( n+1 \right)^{4}+5 \right]}$
Mà: $\frac{n^{4}-\left( n+1 \right)^{4}}{\left[ 2\left( n+1 \right)^{4}+5 \right]\left[ 2\left( n+1 \right)^{4}+5 \right]}<0; \forall n\in N*$
Nên: $\left( u_n \right)$ tăng $\Leftrightarrow u_{n+1}-u_n>0; \forall n\in N*\Leftrightarrow 4-5a<0\Leftrightarrow a>\frac{4}{5}$