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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\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
\(\left(1+1\frac{1}{4}+1\frac{1}{2}+1\frac{3}{4}+2+2\frac{1}{4}+2\frac{1}{2}+2\frac{3}{4}+...+4\frac{3}{4}\right):23\)
= \(\left(\frac{2}{2}+\frac{5}{4}+\frac{3}{2}+\frac{7}{4}+\frac{4}{2}+\frac{9}{4}+\frac{5}{2}+\frac{11}{4}+...+\frac{19}{4}\right):23\)
= \(\left(\frac{4}{4}+\frac{5}{4}+\frac{6}{4}+\frac{7}{4}+\frac{8}{4}+\frac{9}{4}+\frac{10}{4}+\frac{11}{4}+...+\frac{19}{4}\right):23\)
= \(\left(\frac{4+5+6+7+8+9+10+11+...+19}{4}\right):23\)
= \(\left(\frac{\left(19-4\right):1+1x\left(19+4\right):2}{4}\right):23\)
= \(46:23\)
= \(2\)
1/1+2 + 1/1+2+3 +1/1+2+3+4 +...+1/1+2+3+...+50
Ta có 2/2(1+2)+2/2(1+2+3)+...+2/2(1+2+...+50)
=2/6+2/12+2/20+...+2/2550
=2/2.3+2/3.4+...+2/50.51
=2(1/2.3+1/3.4+...+1/50.51)
=2(1/1-1/2+1/2-...+1/50-1/51)
=2.(1-1/51)
=2.50/51=100/51
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