sửa đề :
\(C=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{105}\right)\)
\(C=\left(1+1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\right)\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{105}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{210}\)
\(A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{210}\right)\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)
\(A=2.\frac{13}{30}\)
\(A=\frac{13}{15}\)
Thay A vào ta được :
B = \(\left(1+1+1+1+...+1\right)-\frac{13}{15}\)
B = \(14-\frac{13}{15}\)( có 14 số 1 )
B = \(\frac{197}{15}\)
