cộng hết tất cả 1/1+2+3+.....+10 thì ta chỉ cần cộng 1+2+3+4+5+6+7+8+9+10 là xong rồi tự tính

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.............+\frac{1}{1+2+3+......+10}\)

= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..............+\frac{1}{45}\)

Đến đây bạn làm tiếp nhé

Thua k câu hỏi trước của mình nhé

k là k đánh lộn

Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)

         = \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)

         = \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)

         = \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)

         = 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)

         = \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+10}\)

\(=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)

\(=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=2\cdot\frac{9}{22}\)

\(=\frac{9}{11}\)

Đây mà toán lớp 5 à.

Áp dụng công thức

\(\frac{1}{1+2+...+n}=\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2}{n\left(n+1\right)}\)  ta được

\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+....+50}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)

Ta có : \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+......+50}\)

\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+......+\frac{1}{\frac{50.51}{2}}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{50.51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)

\(=2.\frac{1}{2}-2.\frac{1}{51}\)

\(=1-\frac{2}{51}=\frac{49}{51}\)