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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}\)
= \(1-\frac{1}{20}\)
= \(\frac{19}{20}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+.......+\frac{2}{18.19}+\frac{2}{19.20}.\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=\frac{2.19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...-\frac{1}{19}+\frac{1}{20}\)
\(=1+\frac{1}{20}\)
\(=\frac{1}{20}\)
\(\frac{1}{2\times2}+\frac{1}{3\times3}+....+\frac{1}{20\times20}<\frac{1}{6\times6}+\frac{1}{6\times6}+...+\frac{1}{6\times6}=\frac{1}{36}\times19=\frac{19}{36}<\frac{3}{4}\)
Vậy \(\frac{1}{2\times2}+\frac{1}{3\times3}+...+\frac{1}{20\times20}<\frac{3}{4}\)
Ta có:
\(D=1.2+2.3+3.4+4.5+...+99.100\)
\(\Leftrightarrow3D=1.2.\left(3-0\right)+2.3+\left(4-1\right)+3.4+\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(\Leftrightarrow3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(\Leftrightarrow3D=99.100.101\Leftrightarrow D=\frac{99.100.101}{3}=333300\)
\(B=1.3+2.4+3.5+4.6+...+99.101\)
\(\Leftrightarrow B=\left(1.3+3.5+...+99.101\right)+\left(2.4+4.6+...+98.100\right)\)
\(\Leftrightarrow6B=\left(1.3.\left(5-\left(-1\right)\right)+3.5.\left(7-1\right)+...+99.101.\left(103-97\right)\right)+\left(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\right)\)
\(\Leftrightarrow B=\frac{99.101.103+3}{6}+\frac{98.100.102}{6}=338250\)
Vì các bước gần tương tự như bài a) nên mình bỏ bước.
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
1/2-1/3+1/3-1/4+...+1/9-1/10
=1/2-1/10
=2/5
Chúc bạn học giỏi và thông minh hơn!
= \(\frac{1x1x1}{1x2x4}x\frac{2.2.1}{1.1.2.2}=\frac{1}{8}.1=\frac{1}{8}\)
=1X2X3/1X2X3X4X2= 1/8 =3X2X2X2X5/3X2X2X5X2= 1/1
=1/8X1/1=1/8
Lời giải :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ko chép lại đề :
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ......... + \(\frac{1}{98}\)- \(\frac{1}{99}\)+ \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Cho biểu thức A= 11×2×3 + 12×3×4 + 13×
4×5 +...+ 118×19×20 . So sánh A với 14 .
cố lên mà k