\(B=\left(1+\frac{1}{1.3}\right)+\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...+\left(1+\frac{1}{98.100}\right)\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\right)\)( 98 số 1 ở tồng đầu tiên)
\(=98+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.101}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=98+\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{3}{97.101}\right)+\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=98+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{98}-\frac{1}{100}\right)\)\(=98+\frac{1}{2}.\left(1-\frac{1}{101}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=98+\frac{1}{2}.\frac{100}{101}+\frac{1}{2}.\frac{49}{100}\)
\(=98+\frac{51}{101}+\frac{49}{200}\)
Suy ra phàn nguyên của B là 98.
Vậy phân fnguyên của B là 98.
mình nhầm. bạn thay các chỗ có "97.101" thành "99.101" nhé!
