ta có:
1.2-1/2!+2.3-1/3!+3.4-1/4!+...+99.100-1/100!
=1.2/2!-1/2!+2.3/3!-13!+...+99.100-1/100!
=(1.2/2!+2.3/3!+3.4-4!+...+99.100/100!)-(1/2!+1/3!+...+1/100!)
=(1+1+1/2+...+1/98!)_(1/2!+1/3!+...+1/100!)
=2-1/99!-1/100!<2
Ta xét :
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(=1+1-\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\RightarrowĐPCM\)
Ta có:
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(=1+1-\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{99}-\frac{1}{100}< 2\)
Vậy \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
