Tính :
\(A=1\cdot2+2\cdot3+.......+99\cdot100\)
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Những câu hỏi liên quan
HL
1
30 tháng 8 2020
\(1.2^2+2.3^2+...+99.100^2\)
\(=1.2\left(3-1\right)+2.3\left(4-1\right)+...+99.100\left(101-1\right)\)
\(=1.2.3-1.2+2.3.4-2.3+...+99.100.101-99.100\)
\(=\left(1.2.3+2.3.4+...+99.100.101\right)\)\(-\left(1.2+2.3+...+99.100\right)\)
Chúc học tốt
QD
19 tháng 2 2017
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
25 tháng 10 2021
\(0,125.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}\left(\dfrac{3}{7}-\dfrac{11}{7}\right)=\dfrac{1}{8}.-\dfrac{8}{7}=-\dfrac{1}{7}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
11 tháng 7 2016
1/1.2 + 1/2.3 + 1/3.4 +......+1/99.100
= 1/1 + -1/2 + 1/2 + -1/3 + 1/3 + -1/4 +1/4 +.....+ -1/99 + 1/99 + -1/100
= [ ( -1/2 +1/2) +( -1/3+1/3) + (-1/4 + 1/4) +..... +( -1/99+1/99 ) ] + ( 1/1 + -1/100 )
= 0 + 99/100
= 99/100
11 tháng 7 2016
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
NM
1
20 tháng 7 2016
Gọi \(A=1×2+2×3+..+99×100\)
\(3A=1.2.3+2.3.3+...+999.100.3=1.2\left(3-0\right)+2.3\left(4-1\right)+...+98.99\left(100-97\right)=1.2.3+2.3.4-1.2.3+...-98.99.100-99.100.101=99.100.101\)
\(A=\frac{99.100.101}{3}=333300\)
NT
0
PP
5
1 tháng 8 2018
\(B=1.2+2.3+....+99.100\)
\(\Rightarrow3B=1.2.3+2.3.4+...+99.100.3\)
\(\Rightarrow3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+....+99.100.\left(101-98\right)\)
\(=\left(1.2.3+2.3.4+....+99.100.101\right)-\left(0.1.2+1.2.3+...+98.99.100\right)\)
\(=99.100.101-0.1.2\)
= 999900 - 0
=> B = 999900 : 3 = 333300
Vậy B = 333300
1 tháng 8 2018
B = 1.2 + 2.3 + 3.4 + ...+ 99.100
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ...+99.100.3
3B = 1.2.3 + 2.3.(4-1) + ...+ 99.100.(101-98)
3B = 1.2.3 + 2.3.4 - 1.2.3 + ...+ 99.100.101 - 98.99.100
3B = (1.2.3+2.3.4+...+99.100.101) - (1.2.3+...+98.99.100)
3B = 99.100.101
\(\Rightarrow B=\frac{99.100.101}{3}=333300\)
HN
7
4 tháng 4 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=3.\left(1.2+2.3+3.4+...+99.100\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Rightarrow3A=99.100.101\)
\(\Rightarrow A=\frac{99.100.101}{3}\)
Ta có: A= \(1.2+2.3+3.4+....+99.100\)
=> \(3A=1.2.3+2.3.3+3.4.3+....+99.100.3\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+99.100.101-98.99.100\)
\(\Rightarrow3A=99.100.101\)
\(\Rightarrow3A=999900\)
\(\Rightarrow A=333300\)