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22 tháng 8 2018
3C = 1x2x3 + 2x3x(4-1) + ... + 11x12x(13-9)
3C = 1x2x3 + 2x3x4 - 1x2x3 + ... + 11x12x13 - 9x11x12
3C = 9x11x12
3C = 1188
C = 396
2 tháng 7 2016
\(\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+....+\frac{5}{49.50}\)
\(=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+.....+\frac{1}{49.50}\right)\)
\(=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(=5.\left(\frac{1}{10}-\frac{1}{50}\right)\)
\(=5.\frac{2}{25}\)
\(=\frac{2}{5}\)
2 tháng 7 2016
5/10 x 11 + 5/11 x 12 + 5/12 x 13 + .... + 5/49 x 50
= 5/10 - 5/11 + 5/11 - 5/12 + 5/12 - 5/13 + ....... + 5/49 - 5/50
= 5/10 - 5/50
= 2/5
PT
0
ZV
2
17 tháng 6 2016
A=1x2+2x3+3x4+4x5+......+99x100+100x101
3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+99x100x(101-98)+100x101x(102-99)
3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+...+99x100x101-98x99x100+100x101x102-99x100x101
3A=(1x2x3+2x3x4+3x4x5+4x5x6+...+99x100x101+100x101x102)-(0x1x2+1x2x3+2x3x4+3x4x5+...+98x99x100+99x100x101)
3A=100x101x102
A=100x101x102:3
A=343400
17 tháng 6 2016
A = 1x2 + 2x3 + 3x4 + 4x5 + ... + 99x100 + 100x101
3A = 1x2x(3-0) + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) + 100x101x(102-99)
3A = 1x2x3 - 0x1x2 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100 + 100x101x102 - 99x100x101
3A = 100x101x102 - 0x1x2
3A = 100x101x102
A = 100x101x34
A = 343400
14 tháng 1 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 5 2023
A = \(\dfrac{4}{8\times9}\)+\(\dfrac{4}{9\times10}\)+\(\dfrac{4}{10\times11}\)+...+\(\dfrac{4}{66\times67}\)
A = 4 \(\times\)( \(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+...+\(\dfrac{1}{66\times67}\))
A = 4\(\times\)(\(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\)...+\(\dfrac{1}{66}-\dfrac{1}{67}\))
A = 4\(\times\)(\(\dfrac{1}{8}\) - \(\dfrac{1}{67}\))
A = 4 \(\times\) \(\dfrac{59}{536}\)
A = \(\dfrac{59}{134}\)
A =
DP
26 tháng 6 2017
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
ML
26 tháng 6 2017
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
MN
1
24 tháng 9 2023
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)