Tính tổng:
S= 1/2 + 1/6 + 1/12 + ...+ 1/90
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Những câu hỏi liên quan
8 tháng 4 2015
=1/2 + (1/2*3+1/3*4) + (1/4*5+1/5*6) + (1/6*7+1/7*8) + (1/8*9+1/9*10)
=1/2 + 1/2*3.(1+1/2) + 1/2*5.(1/2+1/3) + 1/2*7.(1/3+1/4) + 1/2*9.(1/4+1/5)
=1/2 + 1/2*3.(3/2) + 1/2*5.(5/6) + 1/2*7.(7/12) + 1/2*9.(9/20)
=1/2 + 1/4 + 1/12 + 1/24 + 1/40
=9/10
NP
8 tháng 4 2015
6 = 2 x 2 + 2 = 2(2+1)
12 = 3 x 3 + 3 = 3(3+1)
20 = 4 x 4 + 4 = 4(4+1)
...
90 = 9 x 9 + 9 = 9(9+1)
Ta có đồng nhất thức sau
1/[n(n+1)] = 1/n - 1/(n+1)
Vậy
1/6 = 1/2 - 1/3
1/12 = 1/3 - 1/4
1/20 = 1/4 - 1/5
....
1/90 = 1/9 - 1/110
S = 1/2 -1/3+1/3-1/4+..............+1/9 -1/10
Tổng là 1/2 + 1/2 - 1/10 = 1 - 1/10 = 9/10
TL
1
XO
7 tháng 6 2019
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
= \(\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)8 số hạng 1
= \(\left(1.8\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
= \(8-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(8-\left(1-\frac{1}{10}\right)\)
= \(8-\frac{9}{10}\)
= \(\frac{71}{10}\)
NT
4
9 tháng 6 2017
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+......+\(\frac{1}{9.10}\)
A=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+........+\(\frac{1}{9}\)-\(\frac{1}{10}\)
A=1-\(\frac{1}{10}\)=9/10
19 tháng 3 2020
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
TD
5
HH
1
24 tháng 5 2016
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}\)
\(=\frac{9}{10}\)
HM
2
HP
13 tháng 12 2015
\(=>S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}+\frac{1}{49.50}\)
\(=>S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
vậy S=49/50
NH
2
6 tháng 8 2016
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
TN
6 tháng 8 2016
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
LH
31 tháng 7 2017
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S=1-\frac{1}{10}\)
\(S=\frac{9}{10}\)
31 tháng 7 2017
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S=1-\frac{1}{10}\)
\(S=\frac{9}{10}\)
S= 1/2 + 1/6 + 1/12 + ...+ 1/90
S=1/1.2 + 1/2.3 +1/3.4+...+1/9.10
S=1-1/2 + 1/2-1/3 + 1/3-1/4+...+1/9-1/10
S=(1-1/10)+(1/2-1/2)+(1/3-1/3)+...+(1/9-1/9)
S=9/10+0+0+...+0
S=9/10Chúc bn học tốt =))