A=100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100
A/100=1/1 x 2 + 1/2 x 3 + 1/3 x 4 +...+1/99 x 100
A/100=2-1/1x2 + 3-2/2x3 + ... + 100-99/99x100
A/100=1-1/2 + 1/2-1/3+...+1/99-1/100
A/100=1-1/100
A/100=99/100
A=99/100x100=99
Vậy A=99.
Ta có:
\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)
\(\Rightarrow100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{100}\right)\Leftrightarrow100.\frac{99}{100}=99\)
\(\text{Ta có :}\)
\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)
\(=100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(100.\left(\frac{1}{1}-\frac{1}{100}\right)=100.\frac{99}{100}=99\)
\(\frac{100}{1.2}\)\(+\)\(\frac{100}{2.3}\)\(+\)\(\frac{100}{3.4}\)\(+\)\(...+\)\(\frac{100}{99.100}\)
\(=\)\(100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=\)\(100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\)\(100.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=\)\(100.\frac{99}{100}\)
\(=\)\(99\)