1/3 + 1/6 + 1/10 + 1/15 +1/21 +1/28 + 1/36 + 1/45
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TM
13 tháng 4 2015
a) thấy dấu cộng ở trước số 6 thành dấu trừ
b) = 2/ 2 + 2/ 6 + 2/ 12 + 2/ 20 + 2/ 30 + 2/ 42 + 2/ 56 + 2/ 72 + 2/ 90
= 2x ( 1/ 1x2 + 1 / 2x3 + 1/ 3x4 + 1/ 4x5 + 1/ 5x6 + 1/ 6x7 + 1/ 7x8 + 1/ 8x9 + 1/ 9x10 )
= 2x ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +1/5 - 1/6 +.. + 1/8- 1/9 + 1/9 - 1/10 )
=2 x( 1 - 1/10 )
=2 x 9/10 = 18/10 = 9 / 5
PC
1
TG
11 tháng 4 2022
1/2 N=1/2x3 + 1/3x4 +...+1/9x10
1/2 N=1/2-1/3+1/3-1/4+...+1/9-1/10
1/2 N=1/2-1/10=2/5
N=2/5:1/2=4/5
PC
1
2 tháng 7 2023
M=2/6+2/12+...+2/90
=2(1/6+1/12+...+1/90)
=2(1/2-1/3+1/3-1/4+...+1/9-1/10)
=2*4/10=8/10=4/5
AH
Akai Haruma
Giáo viên
9 tháng 11 2021
Lời giải:
$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$
$=1-\frac{1}{9}=\frac{8}{9}$
$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$
12 tháng 9 2015
Coi \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right).\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{45}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{90}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)=1-\frac{1}{5}=\frac{1}{4}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{36}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}\)
\(\Rightarrow A=\frac{2}{5}.2\)
\(\Rightarrow A=\frac{4}{5}\)