DH
1
K
Khách
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
S = 1.4 + 4.7 + 7.10 + 10.13 + ... + 61.64
1.4.9 = 1.4.(7 + 2) = 1.4.7 + 1.4.2
4.7.9 = 4.7.(10 - 1) = 4.7.10 - 1.4.7
7.10.9 = 7.10.(13 - 4) = 7.10.13 - 4.7.10
10.13.9 = 10.13.(16 - 7) = 10.13.16 - 7.10.13
.......................................................................
61.64.9 = 61.64.(67 - 58) = 61.64.67 - 58.61.64
Cộng vế với vế ta có:
1.4.9 + 4.7.9 + 7.10.9 +...+ 61.64.9 = 1.4.2 + 61.64.67
9(1.4 + 4.7 + 7.10+ ...+ 61.64) = 261576
1.4 + 4.7 + 7.10 +...+ 61.64 = 261576 : 9
1.4 + 4.7 + 7.10 + ... + 61.64 = 29064
ND
5
BD
18 tháng 8 2023
\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}+\dfrac{3}{64\cdot67}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\)
\(A=1-\dfrac{1}{67}\) < 1
=> A<1
VN
Võ Ngọc Phương
VIP
18 tháng 8 2023
Ta có:
\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{61.64}+\dfrac{3}{64.67}\)
\(=3.\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\right)\)
\(=3.\left(1-\dfrac{1}{67}\right)\)
\(=3.\dfrac{66}{67}\)
\(=\dfrac{198}{67}\)
Vì \(\dfrac{198}{67}\) có tử lớn hơn mẫu nên \(\dfrac{198}{67}>1\)
Vậy \(A>1\)
3 tháng 8 2016
\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{61.64}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{61}-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\frac{63}{64}\)
\(=\frac{21}{32}\)
3 tháng 8 2016
\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{61.61}\)
\(=2.\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{61.64}\right)\)
\(=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\frac{63}{64}\)
\(=\frac{21}{32}\)
KJ
1
15 tháng 1 2022
\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)
NT
2
13 tháng 3 2016
a) 1x( 1+3) ; 4x( 4+3) ; 7 x ( 7+3) ; 10 x ( 10 +3) ; ,,,,,,,,
b)1720
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
st1 = 1.4 = [ 3.(1-1) + 1].[3.(1-1) + 4]
st2 = 4.7 = [3.(2-1) + 1].[3.(2-1) + 4]
st3 = 7.10 = [3.(3-1) + 1].[3.(3-1) + 4]
..........................................................
stn = [3.(n - 1) + 1].(3.(n - 1) + 4]
stn = (3n - 2).(3n + 1)
số hạng thứ 15 của dãy số trên là:
(3.15 - 2).(3.15 + 1) = 43.46 = 1978
M
2
13 tháng 8 2019
1/1.4+1/4.7+1/7.10+...+1/16.19
=[1/1.4+1/4.7+1/7.10+...+1/16.19] x 3
= 3/1.4+3/4.7+3/7.10+...+3/16.19
= 1-1/4+1/4-1/7+1/7-1/10+....+1/16-1/19
=1-1/19
=18/19 :3
=6/19
ĐÂY BẠN NHÉ CHÚC BẠN HỌC TỐT NHỚ K CHO MÌNH
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
A = \(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\)
A= \(\dfrac{2}{2}.\left(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\right)\)
A= \(2.\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{240}\right)\)
A= \(2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
A=\(2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
A=\(2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
A=\(\dfrac{2.3}{16}\)
A= \(\dfrac{3}{8}\)
8 tháng 8 2016
a, 1.6,2.7,3.8,...,50.55
Vậy số hạng thứ 50 của dãy là: 50.55=2750
b, Mik chịu thua
PV
3
30 tháng 4 2017
a) \(A=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(\Leftrightarrow A=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}.\dfrac{102}{103}\)
\(\Leftrightarrow\) \(A=\dfrac{170}{103}\)
b) \(B=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}.\dfrac{16}{51}\)
\(B=\dfrac{8}{51}\)
2 tháng 6 2017
A = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
A = \(\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
A = \(\dfrac{5}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{100}+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{100}-\dfrac{1}{100}\right)-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-0-0-...-0-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{103}{103}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\dfrac{102}{103}\)
A = \(\dfrac{170}{103}\)
B = \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
B = \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
B = \(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
B = \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-0-0-...-0-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{17}{51}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\dfrac{16}{51}\)
B = \(\dfrac{8}{51}\)
A = \(\dfrac{15}{1.4}\) + \(\dfrac{15}{4.7}\) + \(\dfrac{15}{7.10}\) + ... + \(\dfrac{15}{61.64}\)
A = \(5.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}...+\dfrac{3}{61.64}\right)\)
A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{61}\) - \(\dfrac{1}{64}\))
A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{64}\))
A = 5. \(\dfrac{63}{64}\)
A = \(\dfrac{315}{64}\)