Các câu hỏi tương tự
Cho : Sn =\(\frac{5}{1.2.3}+\frac{8}{2.3.4}+...+\frac{3n+2}{n.\left(n+1\right).\left(n+2\right)}\)
CMR : S2008 <2
Tính :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
Tính tổng
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
CMR với n thuộc Z, ta có:
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{5}\right).\left(1+\frac{1}{9}\right)...\left(1+\frac{2}{n^2+3n}\right)< 3\)
cho Sn= \(\frac{1}{1\cdot2\cdot3\cdot4}\)+ \(\frac{1}{2\cdot3\cdot4\cdot5}\)+ ... + \(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\)
CMR: 18<\(\frac{1}{S_n}\)<=24
A1.2+2.3+...+nleft(n+1right)Rightarrow3A1.2.3+2.3.3+...+nleft(n+1right)31.2.3+2.3.left(4-1right)+...+nleft(n+1right)left[left(n+2right)-left(n-1right)right]1.2.3+2.3.4-1.2.3+...+nleft(n+1right)left(n+2right)-left(n-1right)nleft(n+1right)nleft(n+1right)left(n+2right)Rightarrow Afrac{nleft(n+1right)left(n+2right)}{3}
Đọc tiếp
\(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right)3\)
\(=1.2.3+2.3.\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Bài 1: CMR
a) A = \(\frac{\left(n+1\right).\left(n+2\right)....\left(2n-1\right).\left(2n\right)}{2^n}\) là số nguyên.
b) B = \(\frac{3.\left(n+1\right).\left(n +2\right)...\left(3n-1\right).3n}{3^n}\)là số nguyên.