Ta xét :
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=2-\frac{1}{99}-\frac{1}{100}\)
Mà \(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\RightarrowĐPCM\)
1.2−12! +2.3−13! +3.4−14! +....+99.100−1100=2 suy ra 1.2−12! +2.3−13! +3.4−14! +....+99.100−1100<2