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Lời giải:
\(C=\lim\limits_{x\to +\infty}\left[x\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x})...(1+\frac{a_n}{x})}-x\right]\)
\(=\lim\limits_{x\to +\infty}x\left[\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}).....(1+\frac{a_n}{x})}-1\right]\)
\(=\lim\limits _{x\to +\infty}\frac{\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}).....(1+\frac{a_n}{x})}-1}{(1+\frac{a_1}{x})(1+\frac{a_2}{x})..(1+\frac{a_n}{x})-1}.\frac{(1+\frac{a_1}{x})(1+\frac{a_2}{x})...(1+\frac{a_n}{x})-1}{\frac{1}{x}}\)
\(=\lim\limits _{x\to +\infty}(A.B)=\lim\limits_{x\to +\infty}A.\lim\limits_{x\to +\infty}B\)
Với $A$. Đặt \(\sqrt[n]{\prod_{i=1}^n (1+\frac{a_i}{x})}=u\). \(x\to +\infty\Rightarrow \frac{a_i}{x}\to 0\Rightarrow 1+\frac{a_i}{x}\to 1\Rightarrow u\to 1\)
\(\lim\limits_{x\to +\infty}A=\lim\limits_{u\to 1}\frac{u-1}{u^n-1}=\lim\limits_{u\to 1}\frac{1}{u^{n-1}+...+1}=\frac{1}{n}\)
Với $B$
\(\lim\limits _{x\to +\infty}B=\lim\limits _{x\to +\infty}\frac{1+\frac{a_1+a_2+..+a_n}{x}+\frac{a_1a_2+a_2a_3+...+a_{n-1}a_n}{x^2}+....-1}{\frac{1}{x}}\)
\(=\lim\limits _{x\to +\infty}\left(a_1+a_2+...+a_n+\frac{a_1a_2+...+a_{n-1}a_n}{x}+...\right)=a_1+a_2+..+a_n\)
Do đó: $C=\frac{a_1+a_2+...+a_n}{n}$
Đáp án C
a.
\(\Leftrightarrow na_{n+2}-na_{n+1}=2\left(n+1\right)a_{n+1}-2\left(n+1\right)a_n\)
\(\Leftrightarrow\dfrac{a_{n+2}-a_{n+1}}{n+1}=2.\dfrac{a_{n+1}-a_n}{n}\)
Đặt \(b_n=\dfrac{a_{n+1}-a_n}{n}\Rightarrow\left\{{}\begin{matrix}b_1=\dfrac{a_2-a_1}{1}=1\\b_{n+1}=2b_n\end{matrix}\right.\) \(\Rightarrow b_n=2^{n-1}\Rightarrow a_{n+1}-a_n=n.2^{n-1}\)
\(\Leftrightarrow a_{n+1}-\left[\dfrac{1}{2}\left(n+1\right)-1\right]2^{n+1}=a_n-\left[\dfrac{1}{2}n-1\right]2^n\)
Đặt \(c_n=a_n-\left[\dfrac{1}{2}n-1\right]2^n\Rightarrow\left\{{}\begin{matrix}c_1=a_1-\left[\dfrac{1}{2}-1\right]2^1=2\\c_{n+1}=c_n=...=c_1=2\end{matrix}\right.\)
\(\Rightarrow a_n=\left[\dfrac{1}{2}n-1\right]2^n+2=\left(n-2\right)2^{n-1}+2\)
b.
Câu b này đề sai
Với \(n=1\Rightarrow\sqrt{a_1-1}=0< \dfrac{1\left(1+1\right)}{2}\)
Với \(n=2\Rightarrow\sqrt{a_1-1}+\sqrt{a_2-1}=0+1< \dfrac{2\left(2+1\right)}{2}\)
Có lẽ đề đúng phải là: \(\sqrt{a_1-1}+\sqrt{a_2-1}+...+\sqrt{a_n-1}\ge\dfrac{n\left(n-1\right)}{2}\)
Ta sẽ chứng minh: \(\sqrt{a_n-1}\ge n-1\) ; \(\forall n\in Z^+\)
Hay: \(\sqrt{\left(n-2\right)2^{n-1}+1}\ge n-1\)
\(\Leftrightarrow\left(n-2\right)2^{n-1}+2n\ge n^2\)
- Với \(n=1\Rightarrow-1+2\ge1^2\) (đúng)
- Với \(n=2\Rightarrow0+4\ge2^2\) (đúng)
- Giả sử BĐT đúng với \(n=k\ge2\) hay \(\left(k-2\right)2^{k-1}+2k\ge k^2\)
Ta cần chứng minh: \(\left(k-1\right)2^k+2\left(k+1\right)\ge\left(k+1\right)^2\)
\(\Leftrightarrow\left(k-1\right)2^k+1\ge k^2\)
Thật vậy: \(\left(k-1\right)2^k+1=2\left(k-2\right)2^{k-1}+2^k+1\ge2k^2-4k+2^k+1\)
\(\ge2k^2-4k+5=k^2+\left(k-2\right)^2+1>k^2\) (đpcm)
Do đó:
\(\sqrt{a_1-1}+\sqrt{a_2-1}+...+\sqrt{a_n-1}>0+1+...+n-1=\dfrac{n\left(n-1\right)}{2}\)
Ta có:
\({a_1} = 0;{a_2} = 1;{a_3} = 2;{a_4} = 3;{a_5} = 4\).
\({b_1} = 2.1 = 2;{b_2} = 2.2 = 4;{b_3} = 2.3 = 6;{b_4} = 2.4 = 8\).
\({c_1} = 1;{c_2} = {c_1} + 1 = 1 + 1 = 2;{c_3} = {c_2} + 1 = 2 + 1 = 3;{c_4} = {c_3} + 1 = 3 + 1 = 4\).
+ Chu vi đường tròn có bán kính \(n\) là \({d_n} = 2\pi n\).
Ta có: \({d_1} = 2\pi .1 = 2\pi ;{d_2} = 2\pi .2 = 4\pi ;{d_3} = 2\pi .3 = 6\pi ;{d_4} = 2\pi .4 = 8\pi \).
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)
= hệ số xn-1 trên tử/hệ số xn-1 dưới mẫu = \(\dfrac{a_1+a_2+...+a_n}{n}\)
\(\Leftrightarrow n\left(a_{n+2}-a_{n+1}\right)=\left(n+1\right)\left(a_{n+1}-a_n\right)+3n\left(n+1\right)\)
\(\Leftrightarrow\dfrac{a_{n+2}-a_{n+1}}{n+1}=\dfrac{a_{n+1}-a_n}{n}+3\)
Đặt \(\dfrac{a_{n+1}-a_n}{n}=b_n\Rightarrow\left\{{}\begin{matrix}b_1=\dfrac{a_2-a_1}{1}=-6\\b_{n+1}=b_n+3\end{matrix}\right.\)
\(\Rightarrow b_n\) là cấp số cộng với công sai 3
\(\Rightarrow b_n=b_1+\left(n-1\right)d=-6+3\left(n-1\right)=3n-9\)
\(\Rightarrow a_{n+1}-a_n=n\left(3n-9\right)=3n^2-9n\)
\(\Rightarrow a_{n+1}-\left(n+1\right)^3+6\left(n+1\right)^2-5\left(n+1\right)=a_n-n^3+6n^2-5n\)
Đặt \(a_n-n^3+6n^2-5n=c_n\Rightarrow\left\{{}\begin{matrix}c_1=6-1+6-5=6\\c_{n+1}=c_n=...=c_1=6\end{matrix}\right.\)
\(\Rightarrow a_n=n^3-6n^2+5n+6\)
a) Ta có:
\(\begin{array}{l}\left( P \right)\parallel \left( {{A_1}{A_2}{A_3}...{A_6}} \right)\\ \Rightarrow {A_1}^\prime {A_2}^\prime \parallel {A_1}{A_2},{A_2}^\prime {A_3}^\prime \parallel {A_2}{A_3},{A_3}^\prime {A_4}^\prime \parallel {A_3}{A_4},{A_4}^\prime {A_5}^\prime \parallel {A_4}{A_5},{A_5}^\prime {A_6}^\prime \parallel {A_5}{A_6},{A_6}^\prime {A_1}^\prime \parallel {A_6}{A_1}\\ \Rightarrow \frac{{{A_1}^\prime {A_2}^\prime }}{{{A_1}{A_2}}} = \frac{{{A_2}^\prime {A_3}^\prime }}{{{A_2}{A_3}}} = \frac{{{A_3}^\prime {A_4}^\prime }}{{{A_3}{A_4}}} = \frac{{{A_4}^\prime {A_5}^\prime }}{{{A_4}{A_5}}} = \frac{{{A_5}^\prime {A_6}^\prime }}{{{A_5}{A_6}}} = \frac{{{A_6}^\prime {A_1}^\prime }}{{{A_6}{A_1}}}\end{array}\)
Mà \({A_1}{A_2} = {A_2}{A_3} = {A_3}{A_4} = {A_4}{A_5} = {A_5}{A_6} = {A_6}{A_1}\)
\( \Rightarrow {A_1}^\prime {A_2}^\prime = {A_2}^\prime {A_3}^\prime = {A_3}^\prime {A_4}^\prime = {A_4}^\prime {A_5}^\prime = {A_5}^\prime {A_6}^\prime = {A_6}^\prime {A_1}^\prime \)
Vậy đa giác \({A_1}^\prime {A_2}^\prime {A_3}^\prime ...{A_6}^\prime \) là lục giác đều.
b) Ta có:
\(\left. \begin{array}{l}O' \in {A_1}^\prime {A_4}^\prime \subset \left( {S{A_1}{A_4}} \right)\\O' \in {A_3}^\prime {A_6}^\prime \subset \left( {S{A_3}{A_6}} \right)\\\left( {S{A_1}{A_4}} \right) \cap \left( {S{A_3}{A_6}} \right) = SO\end{array} \right\} \Rightarrow O' \in SO\)
Mà \(S.{A_1}{A_2}{A_3}...{A_6}\) là hình chóp đều \( \Rightarrow SO \bot \left( {{A_1}{A_2}{A_3}...{A_6}} \right)\)
Vậy \(OO' \bot \left( {{A_1}{A_2}{A_3}...{A_6}} \right)\)
