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TT
1
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
Đặt biểu thức trên là $A$.
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38.39}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)
\(\Rightarrow A=\frac{185}{741}\)
1
OY
26 tháng 9 2021
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
SC
2
5 tháng 2 2018
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.......+\dfrac{1}{37.38.39}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{37.38}-\dfrac{1}{38.39}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)
\(=\dfrac{370}{741}\)
5 tháng 2 2018
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+......+\dfrac{1}{37.38.39}\)
Ta có:
\(\dfrac{1}{1.2.3}=\dfrac{1}{1.2}-\dfrac{1}{2.3}\); \(\dfrac{1}{2.3.4}=\dfrac{1}{2.3}-\dfrac{1}{3.4}\);.......
\(\Rightarrow A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...........+\dfrac{1}{37.38}-\dfrac{1}{38.39}\)
\(\Rightarrow A=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)
\(=\dfrac{370}{741}\)
Vậy \(A=\dfrac{370}{741}\)
CC
3
LN
20 tháng 9 2017
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)+1\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+2n+n+2}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2}\right)\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2\left(n^2+3n+2\right)}\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2n^2+6n+4}\)
\(S_n=\dfrac{2n^2+6n+4}{4\left(2n^2+6n+4\right)}-\dfrac{4}{4\left(2n^2+6n+4\right)}\)
\(S_n=\dfrac{2n^2+6n+4}{8n^2+48n+16}-\dfrac{4}{8n^2+48n+16}\)
\(S_n=\dfrac{2n^2+6n}{8n^2+48n+16}\)
\(S_n=\dfrac{2\left(n^2+3n\right)}{2\left(4n^2+24n+8\right)}=\dfrac{n^2+3n}{4n^2+24n+8}\)
20 tháng 9 2017
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\\ =>S_n=\dfrac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)
Giải sai r nhéLinh Nguyễn
NN
2
14 tháng 11 2017
theo mình thì
Ta chứng minh được bài toán tổng quát sau
2/[(n-1)n(n+1)] = 1/[(n-1)n] - 1/[n(n+1)]
Áp dụng:
ta có 2C = 1/(1.2) - 1/ (2.3) +1/(2.3) - 1/(3.4) + ...+ 1/18.19 - 1/19.20
= 1/(1.2) - 1/(19.20) = [190 - 1] / 19.20 = 189/380
=> C = 189/ 760
31 tháng 10 2022
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
NT
26 tháng 9 2017
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{48\cdot49\cdot50}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-...+\dfrac{1}{48\cdot49}-\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2450}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{612}{1225}\)
\(=\dfrac{306}{1225}\)