Bài 5: Ôn tập chương Dãy số. Cấp số cộng và cấp số nhân.

bn là giáo viên à

1.

\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{x^2-3x+2\sqrt{x-1}}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2-3x+2+2\left(\sqrt{x-1}-1\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-1\right)\left(x-2\right)+\dfrac{2\left(x-2\right)}{\sqrt{x-1}+1}}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(x-1+\dfrac{2}{\sqrt{x-1}+1}\right)=2\)

\(f\left(2\right)=m-2\)

Hàm liên tục tại \(x=2\) khi \(\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\)

\(\Leftrightarrow m-2=2\Rightarrow m=4\)

thi à

Đề học sinh giỏi, nâu giúp

\(x^2+y^2+xy+2-3x-3y=0\)

\(\Leftrightarrow\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2=1\)

Đặt \(\left\{{}\begin{matrix}x+\dfrac{y}{2}-\dfrac{3}{2}=sina\\\dfrac{\sqrt{3}}{2}\left(y-1\right)=cosa\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=sina-\dfrac{y-3}{2}\\y=\dfrac{2}{\sqrt{3}}cosa+1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=sina-\dfrac{1}{\sqrt{3}}cosa+1\\y=\dfrac{2}{\sqrt{3}}cosa+1\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{3sina+\dfrac{1}{\sqrt{3}}cosa+6}{sina+\dfrac{1}{\sqrt{3}}cosa+8}\)

\(\Rightarrow P.sina-\dfrac{P}{\sqrt{3}}cosa+8P=3sina+\dfrac{1}{\sqrt{3}}cosa+6\)

\(\Rightarrow\left(P-3\right)sina-\left(\dfrac{P+1}{\sqrt{3}}\right)cosa=6-8P\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(P-3\right)^2+\left(\dfrac{P+1}{\sqrt{3}}\right)^3\ge\left(6-8P\right)^2\)

\(\Leftrightarrow47P^2-67P+20\le0\)

\(\Rightarrow\dfrac{20}{47}\le P\le1\)

D. a=0

lim\(\dfrac{an^3+n^2+1}{n^2+n}\)

\(lim\dfrac{an+1+\dfrac{1}{n^2}}{1+\dfrac{1}{n}}=an+1=1\)

\(\Rightarrow a=0\)

1. Áp dụng công thức tổng cấp số nhân:

\(S_n=u_1.\dfrac{q^n-1}{q-1}=2.\dfrac{2^n-1}{2-1}=2.\left(2^n-1\right)=2^{n+1}-2\)

2. \(\left\{{}\begin{matrix}u_2+u_5=12\\u_4+u_8=22\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(u_1+d\right)+\left(u_1+4d\right)=12\\\left(u_1+3d\right)+\left(u_1+7d\right)=22\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u_1+5d=12\\2u_1+10d=22\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=1\\d=2\end{matrix}\right.\)

\(\Rightarrow u_n=u_1+\left(n-1\right)d=1+\left(n-1\right)2=2n-1\)

\(\Rightarrow S_n=\dfrac{n\left(u_1+u_n\right)}{2}=\dfrac{n\left(1+2n-1\right)}{2}=n^2\)

3. \(\left\{{}\begin{matrix}u_1+u_2=4\\u_4+u_1=28\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_1q=4\\u_1q^3+u_1=28\end{matrix}\right.\)

\(\Rightarrow\dfrac{q^3+1}{q+1}=\dfrac{28}{4}\Rightarrow q^2-q+1=7\)

\(\Rightarrow q^2-q-6=0\Rightarrow\left[{}\begin{matrix}q=3\\q=-2\end{matrix}\right.\)

thì ko hoặc có

Do n lẻ, đặt \(n=2m+1\)

\(\Rightarrow S=C_{2m+1}^1+C_{2m+1}^2+...+C_{2m+1}^m\)

Áp dụng đẳng thức: \(C_n^k=C_n^{n-k}\)

\(\Rightarrow S=C_{2m+1}^{2m}+C_{2m+1}^{2m-1}+...+C_{2m+1}^{m+1}\)

\(\Rightarrow2S=S+S=C_{2m+1}^1+C_{2m+1}^2+...+C_{2m+1}^{2m}\)

\(=C_{2m+1}^0+C_{2m+1}^1+...+C_{2m+1}^{2m+1}-\left(C_{2m+1}^0+C_{2m+1}^{2m+1}\right)\)

\(=2^{2m+1}-2\)

\(\Rightarrow S=2^{2m}-1\) luôn lẻ (đpcm)

Do \(\left(u_n\right)\) là dãy tăng nên \(u_n\ge2018,\forall n\ge1\)

Ta có : \(u^2_n+2018u_n-2020u_{n+1}=0\) \(\Leftrightarrow u_{n+1}=\dfrac{u^2_n+2018u_n+1}{2020}\)

\(\Leftrightarrow u_{n+1}-1=\dfrac{u^2_n+2018u_n-2019}{2020}\Leftrightarrow2020\left(u_{n+1}-1\right)=\left(u_n-1\right)\left(u_n+2019\right)\)

\(\Leftrightarrow\dfrac{2020}{\left(u_n-1\right)\left(u_n+2019\right)}=\dfrac{1}{u_{n+1}-1}\Rightarrow\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\left(1\right)\)

Thay n bởi 1 , 2, 3 , ..... , n vào (1) và cộng vế với vế các đẳng thức ta suy ra:

\(S_n=\dfrac{1}{u_1+2019}+\dfrac{1}{u_2+2019}+...+\dfrac{1}{u_n+2019}\)

\(=\dfrac{1}{u_1-1}-\dfrac{1}{u_{n+1}-1}=\dfrac{1}{2018}-\dfrac{1}{u_{n+1}-1}\)

Do \(\left(u_n\right)\) là dãy số tăng nên có 2 trường hợp xảy ra:

th1:

Dãy \(\left(u_n\right)\) bị chặn trên suy ra tồn tại \(lim\) \(u_n\) . Giả sử lim \(u_n\) = x thì \(x\ge2018\)

chuyển qua giới hạn hệ thức \(\left(1\right)khi\) \(n\rightarrow+\infty\) , ta có:

\(x^2+2018x-2020x+1=0\)

\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\) ( điều này vô lý)

Dãy \(\left(u_n\right)\) ko bị chặn trên , do \(\left(u_n\right)\) tăng và ko bị chặn nên:

\(lim\) \(u_n\)\(=+\infty\Rightarrow lim\left(u_{n+1}-1\right)=+\infty\Rightarrow lim\dfrac{1}{u_{n+1}-1}=0\)

Do vậy: \(limS_n=lim\left(\dfrac{1}{2018}-\dfrac{1}{u_{n+1}-1}\right)=\dfrac{1}{2018}\)

á cái này bt nek mak gõ hơi chậm nha nói trc cỡ 15p ms xog ó:>

??? Câu hỏi đâu mà giúp

\(u_{n+1}=\dfrac{n\left(u_n+2\right)+n^2+1}{n+1}\)

\(\Rightarrow\left(n+1\right)u_{n+1}=nu_n+n^2+2n+1\)

\(\Rightarrow\left(n+1\right)u_{n+1}-\dfrac{1}{3}\left(n+1\right)^3-\dfrac{1}{2}\left(n+1\right)^2-\dfrac{1}{6}\left(n+1\right)=n.u_n-\dfrac{1}{3}n^3-\dfrac{1}{2}n^2-\dfrac{1}{6}n\)

Đặt \(v_n=u.u_n-\dfrac{1}{3}n^3-\dfrac{1}{2}n^2-\dfrac{1}{6}n\Rightarrow\left\{{}\begin{matrix}v_1=1-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{6}=0\\v_{n+1}=v_n=...=v_1=0\end{matrix}\right.\)

\(\Rightarrow n.u_n-\dfrac{1}{3}n^3-\dfrac{1}{2}n^2-\dfrac{1}{6}n=0\)

\(\Rightarrow u_n=\dfrac{1}{3}n^2+\dfrac{1}{2}n+\dfrac{1}{6}=\dfrac{\left(n+1\right)\left(2n+1\right)}{6}\)