Câu hỏi của GT 6916 - Toán lớp 7 - Học toán với OnlineMath
Bạn tham khảo.
Câu hỏi của GT 6916 - Toán lớp 7 - Học toán với OnlineMath
Bạn tham khảo.
Tính tổng
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
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
\(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right)3\)
\(=1.2.3+2.3.\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
tính: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{a\left(a+1\right)\left(a+2\right)}\)
Tính tổng \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{a\left(a+1\right)\left(a+2\right)}\)
Cho \(S_n=\frac{5}{1.2.3}+\frac{8}{2.3.4}+...+\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)
\(CMR:\)\(S_{2008}< 2\)
Giúp mk vs ạ
Cảm ơn!!!
tính các tích sau với nEN, n lớn hơn bằng 2
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
b)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
c)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)