G= \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}\)
G= \(\frac{\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}\)
G = \(\frac{\frac{1.2+2.3+...+98.99}{2}}{1.2+2.3+3.4+...+98.99}\)
G= \(\frac{1}{2}\)
Tính A : \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+......+\left(1+2+3+....+98\right)}{1.2+2.3+3.4+.....98.99}\)
tính
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.2+2.3+3.4+...+98.99}\)
Tính
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
B=\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Tính:
a)\(D\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3..+98\right)}{1.98+2.97+...+98.1}\)
b)chứng minh rằng biểu thức E có giá trị bằng \(\frac{1}{2}\)
\(E=\frac{1.98+2.97+...+98.1}{1.2+2.3+...+98.99}\)
Chứng minh: A = 1.2 + 2.3 + 3.4 + 4.5 +.......+ n. (n+1) = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
tính
b=\(\left(\frac{2}{2.3}-1\right).\left(\frac{2}{3.4}-1\right).\left(\frac{2}{4.5}-1\right).....\left(\frac{2}{2008.2009}-1\right)\)
Chứng minh : A = 1.2 + 2.3 + 3.4 + 4.5 + ... + n.(n+1) = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)=?\)
