\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\) \(< \frac{1}{4}\)
Tính:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
a) So \(M=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)vs-\frac{1}{2}\)
b) \(N=\frac{\sqrt{x}+1}{\sqrt{x}-3}\). Tìm \(x\in Z\) để \(N\)là số nguyên dương
Chứng minh vs \(\forall n\) nguyên dương thì \(S_n=1^3+2^3+3^3+...+n^3=\left[\frac{n\left(n+1\right)}{2}\right]^2\)
THU GỌN BIỂU THỨC SAU
\(\left(\frac{n-1}{1}+\frac{n-2}{2}+\frac{n-3}{3}+...+\frac{2}{n-2}+\frac{1}{n-1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}\right)\)
Tìm x : \(\left(1.2.3+2.3.4+...+2015.2016.2017\right)-\left(5.6+6.7+...+99.100\right)\) \(=\left(5+10+15+...+2000\right)-\frac{1}{1.3}-\frac{1}{3.5}-...-\frac{1}{97.99}-4x\)
cho x,y,z là các số thực dương khác 1 và xyz=1. Chứng minh rằng \(\frac{x^2}{\left(x-1\right)^2}+\frac{y^2}{\left(y-1\right)^2}+\frac{z^2}{\left(z-1\right)^2}\ge1\)
Chứng minh với \(\forall n\in N\)* thì \(1^3+2^3+3^3+...+n^3=\left[\frac{n\left(n+1\right)^{ }}{2}\right]^2\)
THU GỌN BIỂU THỨC SAU
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
Rút gọn C = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
