Bộ 24 Đề kiểm tra Giữa kì 2 Toán 11 có đáp án (Mới nhất) (Đề 4)

Ta có  $u_n-5=\frac{6{.5}^n+5{.2}^n}{5^n+2^n}-5=\frac{5^n}{5^n+2^n}$$\Rightarrow \frac{1}{u_n-5}=\frac{5^n+2^n}{5^n}=1+{\left(\frac{2}{5}\right)}^n$

$\Rightarrow S=\frac{1}{u_1-5}+\frac{1}{u_2-5}+...+\frac{1}{u_{2021}-5}=\left(1+{\left(\frac{2}{5}\right)}^1\right)+\left(1+{\left(\frac{2}{5}\right)}^2\right)+...+\left(1+{\left(\frac{2}{5}\right)}^{2021}\right)=\left(2021+\frac{2}{5}.\frac{1-{\left(\frac{2}{5}\right)}^{2021}}{1-\frac{2}{5}}\right)=\left(2021+\frac{2}{3}.\left(1-{\left(\frac{2}{5}\right)}^{2021}\right)\right)=\left(\frac{6065}{3}-\frac{2}{3}{\left(\frac{2}{5}\right)}^{2021}\right).$

Mà $S=\frac{a}{3}-\frac{b}{3}{\left(\frac{2}{5}\right)}^c$  nên $a=6065;$$b=2;$$c=2021$.

Nên $a+2b^2-2c=6065+2{.2}^2-2.2021$$=2031$

+) Ta có: $u_n=2021u_{n-1}-1$$\iff u_n-\frac{1}{2020}=2021\left(u_{n-1}-\frac{1}{2020}\right)$

+) Đặt $v_n=u_n-\frac{1}{2020}$. Ta có  $v_1=u_1-\frac{1}{2020}=-1-\frac{1}{2020}=-\frac{2021}{2020}$ và $v_n=2021v_{n-1},\forall n\ge 2$

Suy ra dãy (v_n) là cấp số nhân với công bội là q = 2021, $v_1=-\frac{2021}{2020}$

Khi đó $v_n=v_1.q^{n-1}=-\frac{2021}{2020}.{2021}^{n-1}$$=-\frac{{2021}^n}{2020},\forall n\ge 1$

Do đó $u_n=v_n+\frac{1}{2020}=-\frac{{2021}^n}{2020}+\frac{1}{2020},$$\forall n=1,2,...$

+) Ta có: $\lim \frac{u_n}{{2021}^n}=\lim \left(-\frac{1}{2020}+\frac{1}{2020{.2021}^n}\right)$$=-\frac{1}{2020}$

Vậy  $\lim \frac{u_n}{{2021}^n}=-\frac{1}{2020}$

a) Chứng minh $\left({\mathrm{AB}}^{\text{'}}{D}^{\text{'}}\right)\text{//}\left({\mathrm{BC}}^{\text{'}}D\right)$.

Từ giả thiết ta có $\left(\begin{array}{c}\mathrm{BD}\text{//}{B}^{\text{'}}{D}^{\text{'}}\\ {\mathrm{AB}}^{\text{'}}\text{//}{\mathrm{DC}}^{\text{'}}\end{array}\right)\Rightarrow \left({\mathrm{AB}}^{\text{'}}{D}^{\text{'}}\right)\text{//}\left({\mathrm{BC}}^{\text{'}}D\right)$

b) Chứng minh ${\mathrm{AC}}^{\text{'}}\perp A^{\text{'}}B$

Ta có: $\overrightarrow{{\mathrm{AC}}^{\text{'}}}.\overrightarrow{A^{\text{'}}B}=\left(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{{\mathrm{AA}}^{\text{'}}}\right).$$\left(\overrightarrow{\mathrm{AB}}-\overrightarrow{{\mathrm{AA}}^{\text{'}}}\right)$

$={\overrightarrow{\mathrm{AB}}}^2+\overrightarrow{\mathrm{AD}}.\overrightarrow{\mathrm{AB}}+\overrightarrow{{\mathrm{AA}}^{\text{'}}}.\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{AB}}.\overrightarrow{{\mathrm{AA}}^{\text{'}}}-\overrightarrow{\mathrm{AD}}.\overrightarrow{{\mathrm{AA}}^{\text{'}}}-{\overrightarrow{{\mathrm{AA}}^{\text{'}}}}^2\Rightarrow \overrightarrow{{\mathrm{AC}}^{\text{'}}}\perp \overrightarrow{A^{\text{'}}B}\Rightarrow {\mathrm{AC}}^{\text{'}}\perp A^{\text{'}}B$.

c) Chứng minh $\mathrm{MN}\text{//}\left({\mathrm{AB}}^{\text{'}}{D}^{\text{'}}\right)$

Dễ thấy M, N không thuộc (AB'D').

$\mathrm{MN}\text{//}\left({\mathrm{AB}}^{\text{'}}{D}^{\text{'}}\right)\iff \overrightarrow{\mathrm{MN}},\overrightarrow{{\mathrm{AB}}^{\text{'}}},\overrightarrow{{\mathrm{AD}}^{\text{'}}}$ đồng phẳng $\iff \exists m,n:\overrightarrow{\mathrm{MN}}=m\overrightarrow{{\mathrm{AB}}^{\text{'}}}+n\overrightarrow{{\mathrm{AD}}^{\text{'}}}$

Đặt $\overrightarrow{\mathrm{AB}}=\overrightarrow{a},\overrightarrow{\mathrm{AD}}=\overrightarrow{b},\overrightarrow{{\mathrm{AA}}^{\text{'}}}=\overrightarrow{c}$

 $\overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{AN}}-\overrightarrow{\mathrm{AM}}=\overrightarrow{{\mathrm{AA}}^{\text{'}}}+\overrightarrow{A^{\text{'}}N}-\frac{1}{5}\overrightarrow{\mathrm{AD}}=\overrightarrow{c}+\frac{2}{5}\overrightarrow{A^{\text{'}}C}-\frac{1}{5}\overrightarrow{b}=\overrightarrow{c}+\frac{2}{5}\left(\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}\right)-\frac{1}{5}\overrightarrow{b}=\frac{2}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}+\frac{3}{5}\overrightarrow{c}$

$\overrightarrow{{\mathrm{AB}}^{\text{'}}}=\overrightarrow{a}+\overrightarrow{c}$, $\overrightarrow{{\mathrm{AD}}^{\text{'}}}=\overrightarrow{b}+\overrightarrow{c}$

$\overrightarrow{\mathrm{MN}}=m\overrightarrow{{\mathrm{AB}}^{\text{'}}}+n\overrightarrow{{\mathrm{AD}}^{\text{'}}}$$\iff \frac{2}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}+\frac{3}{5}\overrightarrow{c}$$=m\overrightarrow{a}+n\overrightarrow{b}+\left(m+n\right)\overrightarrow{c}$$\iff \left(\begin{array}{c}m=\frac{2}{5}\\ n=\frac{1}{5}\end{array}\right)$$\Rightarrow \overrightarrow{\mathrm{MN}}=\frac{2}{5}\overrightarrow{{\mathrm{AB}}^{\text{'}}}+\frac{1}{5}\overrightarrow{{\mathrm{AD}}^{\text{'}}}$

Vậy MN // (AB'D').