tính tổng : 2/1x4 + 2/4.7 + 2/7x10 + ....+ 2/97x100 : giúp mình với
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Những câu hỏi liên quan
DT
1
10 tháng 10 2018
A = 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/97.100
3A = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100 = (4-1)/1.4 + (7-4)/4.7 + (10-7)/7.10 + ... + (100-97)/97.100
= 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100 = 1 - 1/100 = 99/100
=> A = 33/100
A = x/2 => x = 2.A = 33/50
9 tháng 5 2016
\(A=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{97\times100}\right)\)
\(A=3\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3\times\left(1-\frac{1}{100}\right)\)
\(A=3\times\frac{99}{100}\)
\(A=\frac{297}{100}\)
HP
9 tháng 5 2016
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+......+\frac{3^2}{97.100}\)
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Đặt \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
Ta có: \(S=\frac{3}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{97.100}\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)
\(S=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=3.S=3.\frac{99}{100}=\frac{297}{100}\)
NT
0
DN
2
S
8 tháng 7 2019
\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)
DT
2
N
19 tháng 7 2017
\(A=\frac{1}{1.4}+\frac{1}{4.7}+.....+\frac{1}{97.100}\)
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{97.100}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\)
\(3A=1-\frac{1}{100}\)
\(3A=\frac{99}{100}\)
\(A=\frac{99}{100}:3\)
\(A=\frac{33}{100}\)
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)\)
\(=2.\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
_____
b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)
\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)
\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=3.\left(1-\dfrac{1}{101}\right)\)
\(=3.\dfrac{100}{101}\)
\(=\dfrac{300}{101}\)
AH
Akai Haruma
Giáo viên
6 tháng 12 2023
Bài 1:
$M=3.4.5+4.5.6+...+13.14.15$
$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$
$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$
$=-2.3.4.5+13.14.15.16=43560$
$M=43560:4=10890$
AH
Akai Haruma
Giáo viên
6 tháng 12 2023
Bài 2:
a.
$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$
$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$
$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
$M=\frac{99}{100}:3=\frac{33}{100}$
TB
2
M
30 tháng 5 2019
1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100
= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)
= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)
= 1/3(1 - 1/100)
= 1/3*99/100
= 33/100
= 2/3 x ( 3/1x4 + 3/4.7 + 3/7x10 + ....+ 3/97x100)
= 2/3 x (1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
= 2/3 x (1- 1/100)
= 2/3 x 99/100
= 33/50
Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{2}{3}.\frac{99}{100}=\frac{198}{300}=\frac{33}{50}\)