\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{2006.2007}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(=1-\dfrac{1}{2007}=\dfrac{2006}{2007}\)
tính tổng S= (1.2)² + (2.3)² + (3.4)² + … + [n(n + 1)]²
Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
Tính tổng
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Tìm S2 = 1.2+2.3+3.4+...+n.(n+1)
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
tính nhanh
A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
giải đầy đủ mình like
1, Tìm X
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
Chứng minh zới k thuộc N* ta luôn có: k(k+1)(k+2)-(k-1)k(k+1) áp dụng tính tổng: S=1.2+2.3+3.4+..............+n(n+1)
Mình cần gấp!!!
HELP ME!!
