\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}\)
\(=\frac{17}{51}-\frac{1}{51}\)
\(=\frac{16}{51}\)
tính tổng
A=\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{29.30}\)
B=\(\dfrac{4}{7.11}+\dfrac{9}{11.20}+\dfrac{5}{20.25}\)
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...-\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
Ghi cách giải lun nha
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+................+\frac{1}{99.100}\). Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
Cho biểu thức A= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...................+\frac{1}{99.100}\). Chứng minh \(\frac{7}{12}< A< \frac{5}{6}\)
Chứng minh rằng: \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+..................+\frac{99.100-1}{100!}< 2\)
chứng minh rằng
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
1. Chứng minh rằng:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100}< 1\)
2. Chứng minh rằng:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}< 2\)
Chứng minh rằng :\(\dfrac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\dfrac{1}{2}\)
Câu 1 : Tính và so sánh :
C = \(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\) so sánh với \(\dfrac{2n+2}{3n}\)
- Please help me !!!
