Lời giải:
Ta có: \(\frac{1}{k(k+1)(k+2)}=\frac{1}{2}.\frac{2}{k(k+1)(k+2)}=\frac{1}{2}.\frac{(k+2)-k}{k(k+1)(k+2)}\)

\(=\frac{1}{2}\left(\frac{k+2}{k(k+1)(k+2)}-\frac{k}{k(k+1)(k+2)}\right)=\frac{1}{2}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)\)

Áp dụng vào bài toán:

\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)

\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)

.......

\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\cdot\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{n^2+3n+2-2}{2\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

Bài 1 :

Để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì \(x^3+x^2-x-1=0\)

\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy,.........

\(\dfrac{1}{n}-\dfrac{1}{n+k}=\dfrac{n+k}{n\left(n+k\right)}-\dfrac{n}{n\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{k}{n\left(n+k\right)}\)

\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)(đpcm)

Ủa đề bài như này là sao bạn? Cho dãy x(k), nhưng lại đi tìm u(n)?

Ok start

\(\dfrac{1}{2!}=\dfrac{2!-1}{2!}=1-\dfrac{1}{2!};\dfrac{2}{3!}=\dfrac{1}{3}=\dfrac{3!-2!}{3!.2!}=\dfrac{1}{2!}-\dfrac{1}{3!}\)

\(\Rightarrow\dfrac{k}{\left(k+1\right)!}=\dfrac{1}{k!}-\dfrac{1}{\left(k+1\right)!}\)

Explain: \(\dfrac{1}{k!}-\dfrac{1}{\left(k+1\right)!}=\dfrac{\left(k+1\right)k!-k!}{k!\left(k+1\right)!}=\dfrac{k+1-1}{\left(k+1\right)!}=\dfrac{k}{\left(k+1\right)!}\)< Có nên xài quy nạp mạnh cho chặt chẽ hơn ko nhỉ?>

Nhớ lại 1 bài toán lớp 6 cũng có dạng như này

\(\Rightarrow x_k=1-\dfrac{1}{\left(k+1\right)!}\)

Xet \(x_{k+1}-x_k=1-\dfrac{1}{\left(k+2\right)!}-1+\dfrac{1}{\left(k+1\right)!}=\dfrac{1}{\left(k+1\right)!}-\dfrac{1}{\left(k+2\right)!}>0\Rightarrow x_{k+1}>x_k\)

\(\Rightarrow x_1< x_2< ...< x_{2011}\Rightarrow x_1^n< x_2^n< ...< x_{2011}^n\)

\(\Rightarrow\sqrt[n]{x_1^n+x_2^n+...+x_{2011}^n}< \sqrt[n]{x_{2011}^n+x^n_{2011}+...+x^n_{2011}}=\sqrt[n]{2011.x^n_{2011}}=x_{2011}.\sqrt[n]{2011}\)

Mat khac: \(x_{2011}=\sqrt[n]{x^n_{2011}}< \sqrt[n]{x_1^n+x_2^n+...+x_{2011}^n}\)

\(\Rightarrow x_{2011}< \sqrt[n]{x^n_1+x_2^n+...+x_{2011}^n}< \sqrt[n]{2011}x_{2011}\)

\(\lim\limits x_{2011}=1-\dfrac{1}{2012!}\)

\(\lim\limits\sqrt[n]{2011}x_{2011}=\lim\limits2011^0.x_{2011}=1-\dfrac{1}{2012!}\)

\(\Rightarrow\lim\limits\left(u_n\right)=1-\dfrac{1}{2012!}\)

Xin dung cuoc choi tai day, ban check lai xem dung ko, tinh tui hay au co khi sai :v