CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
CMR : \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\forall n\)
Ta có
\(\frac{1}{n-1}\)- \(\frac{1}{n}\)> \(\frac{1}{n^2}\)> \(\frac{1}{n}\)- \(\frac{1}{n+1}\)
chứng tỏ với mọi n\(\in\)N* ta luôn có:\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
áp dụng tính tổng sau:\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
So sánh các cặp phân số sau:
a) \(\frac{n}{n+1}\)và\(\frac{n+2}{n+3}\)\(\forall\)n \(\in\)\(ℕ\)
b) \(\frac{n}{2n+1}\)và \(\frac{2n+3}{4n+2}\)\(\forall\)n \(\in\)\(ℕ\)
c) \(\frac{n}{n+3}\)và\(\frac{2n+1}{3n+4}\)\(\forall\)n\(\inℕ\)
d) \(\frac{2017}{2020}\)và\(\frac{2018}{2019}\)
1.Cho E=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+n}\right)\)và F=\(\frac{n+2}{n}\)\(\forall\)\(n\inℕ^∗\)Tính \(\frac{E}{F}\)
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}(n\in N,n>hoac=2)\)
Chứng tỏ \(A=\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}=\frac{\frac{1}{ }}{2}\times\left(\frac{1}{n\times\left(n+1\right)}-\frac{1}{\left(n+1\right)\times\left(n+2\right)}\right)\)với n\(\in\)N*