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ĐK của pt là \(n\ge2\)
\(\left(1+x\right)^n=C_n^0+x.C_n^1+x^2.C_n^2+x^3.C^3_n+x^4.C_n^4+...+x^n.C_n^n\)
\(\Rightarrow n\left(1+x\right)^{n-1}=C_n^1+2x.C_n^2+3x^2.C^3_n+4x^3.C_n^4...+n.x^{n-1}.C^n_n\) ( đạo hàm hai vế )
\(\Rightarrow n\left(n-1\right)\left(x+1\right)^{n-2}=2.C_n^2+2.3.x.C_n^3+3.4.x^2.C_n^4+...+\left(n-1\right)n.x^{n-2}.C_n^n\) ( đạo hàm hai vế )
Thay x=1 ta được: \(n\left(n-1\right).2^{n-2}=2.C_n^2+2.3.C^3_n+3.4.C_n^4+...+\left(n-1\right).n.C^n_n\)
\(\Leftrightarrow n\left(n-1\right).2^{n-2}=64n.\left(n-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}n\left(n-1\right)=0\\2^{n-2}=64\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=0;n=1\left(ktm\right)\\n=8\left(tm\right)\end{matrix}\right.\)
Vậy \(n=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Với \(n=1\Rightarrow1.2=\frac{1.2.3}{3}\) (đúng)
- Giả sử đúng với \(n=k\) hay \(1.2+...+k\left(k+1\right)=\frac{k\left(k+1\right)\left(k+2\right)}{3}\)
Ta cần chứng minh nó đúng với \(n=k+1\) hay:
\(1.2+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)
Thật vậy:
\(1.2+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)\)
\(=\frac{k\left(k+1\right)\left(k+2\right)}{3}+\left(k+1\right)\left(k+2\right)\)
\(=\left(k+1\right)\left(k+2\right)\left[\frac{k}{3}+1\right]=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\lim u_n=\lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)
b.
\(u_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}\)
\(\Rightarrow\lim u_n=\lim\left(1-\dfrac{1}{n+1}\right)=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a=\lim\limits_{x\rightarrow-3}\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{1}{x-3}=-\dfrac{1}{6}\)
\(b=\lim\limits_{x\rightarrow2}\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{x+3}{x+2}=\dfrac{5}{4}\)
\(c=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+5\right)\left(x-4\right)}=\lim\limits_{x\rightarrow4}\dfrac{x+4}{x+5}=\dfrac{8}{9}\)
\(d=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\dfrac{x+2}{x-1}=4\)
\(e=\lim\limits_{x\rightarrow2}\dfrac{x+7-9}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{x+7}+3}=\dfrac{1}{6}\)
\(f=\lim\limits_{x\rightarrow1}\dfrac{x+3-4}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(h=\lim\limits_{x\rightarrow-3}\dfrac{x+7-4}{\left(x+3\right)\left(\sqrt{x+7}+2\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+3}{\left(x+3\right)\left(\sqrt{x+7}+2\right)}=\lim\limits_{x\rightarrow-3}\dfrac{1}{\sqrt{x+7}+2}=\dfrac{1}{4}\)
Bài 1:
a,
= limx->-3 \(\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}\)
= limx->3 x-3
= -3 -3
= -6
b,
= limx->2 \(\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\)
= limx->2 \(\dfrac{x+3}{x+2}\)
= \(\dfrac{5}{4}\)
c,
= limx->4 \(\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+5\right)}\)
= limx->4 \(\dfrac{\left(x+4\right)}{\left(x+5\right)}\)
= \(\dfrac{8}{9}\)
d,
= limx->2 \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x-1\right)}\)
= limx->2 \(\dfrac{\left(x+2\right)}{\left(x-1\right)}\)
= 4
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(y'=6x^2+6x\Rightarrow y'\left(1\right)=12\)
Đáp án B
2. \(y'=\dfrac{7}{\left(x+3\right)^2}\Rightarrow y'\left(1\right)=\dfrac{7}{16}\) (A)
3. \(y'=8x^3+9x^2-3\Rightarrow y'\left(3\right)=294\)
Tất cả các đáp án đều sai
4. Tiếp tục là 1 câu đề bài sai
Hàm số không xác định tại \(x=1\Rightarrow\) không liên tục tại \(x=1\Rightarrow\) không tồn tại đạo hàm tại \(x=1\)
5.
\(f'\left(x\right)=7x^6+20x^4+6x\)
\(\Rightarrow f'\left(2\right)=780\)
6.
\(y'=\dfrac{3}{\left(x+1\right)^2}\)
\(\Rightarrow y'\left(2\right)=\dfrac{1}{3}\) ; \(y\left(2\right)=1\)
Phương trình tiếp tuyến:
\(y=\dfrac{1}{3}\left(x-2\right)+1\Leftrightarrow y=\dfrac{1}{3}x+\dfrac{1}{3}\)
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\(=\lim\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(=\lim\left(1-\dfrac{1}{n+1}\right)=1\)
\(B=1.2+2.3+3.4+...+99.100\)
\(3B=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)
\(3B=99.100.101\)
\(B=\dfrac{99.100.101}{3}=333300\)
giải rõ nha ribi