\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+\frac{1}{3x4x5}+....+\frac{1}{37x38x39}\)
K
Khách
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Những câu hỏi liên quan
13 tháng 8 2016
\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)
\(2A=\frac{741}{1482}-\frac{1}{1482}\)
\(2A=\frac{370}{741}\)
\(A=\frac{370}{741}:2=\frac{185}{741}\)
12 tháng 8 2016
A = 2/1x2x3 + 2/2x3x4 + 2/3x4x5 + ... + 2/36x37x38 + 2/37x38x39
A = 1/1x2 - 1/2x3 + 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ...+ 1/36x37 - 1/37x38 + 1/37x38 - 1/38x39
A = 1/2 - 1/38x39
A = 370/741
Tớ ko chắc là đúng đâu
NH
5
28 tháng 6 2015
\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(=\frac{1}{1\times2}-\frac{1}{38\times39}=\frac{1}{2}-\frac{1}{1482}=\frac{370}{741}\)
K
28 tháng 6 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{36.37.37}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{36.37.38}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{36.37}-\frac{1}{37.38}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{37.38}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1406}\right)\)
\(=\frac{1}{2}.\frac{351}{703}\)
\(=\frac{351}{1046}\)
NH
3
DP
13 tháng 8 2017
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+......+\frac{1}{48.49.50}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\frac{612}{1225}=\frac{612}{2450}=\frac{306}{1225}\)
T
22 tháng 3 2018
Do not ask why hay quá!
Đặt \(T=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\);\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\);. . . ; \(\frac{1}{48.49}-\frac{1}{49.50}=\frac{1}{48.49.50}\)
Rút ra dạng tổng quát,ta có: (mình nói thêm nhé)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
Ta nhận thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\);.....
\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{49.50}=\frac{612}{1225}\)
\(\Rightarrow T=\frac{612}{\frac{1225}{2}}=\frac{306}{1225}\)
Vậy .. . .
DN
11 tháng 3 2018
\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)
\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)
11 tháng 3 2018
Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)
b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)
\(\Rightarrow2D=1-\frac{1}{3^8}\)
\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)
UI
2
HT
2
NN
28 tháng 1 2021
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.......+\frac{1}{8.9.10}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+......+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.......+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
21 tháng 5 2015
A=\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{37x38x39}\)
=\(\frac{1}{2}x\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{37x38}-\frac{1}{38x39}\right)=\frac{1}{2}x\left(\frac{1}{2}-\frac{1}{38x39}\right)=\frac{185}{741}\)
11 tháng 8 2018
Sửa đề chút:
\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{98x99x100}\)
\(=\frac{1}{2}.\left(\frac{2}{1x2x3}+\frac{2}{2x3x4}+...+\frac{2}{98x99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{98x99}-\frac{1}{99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{4}-\frac{1}{99.200}< 1\)
đpcm