1/3 - 1/3^2 + 1/3^3 - ... + 1/3^99 - 1/3^100
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Gấp ạa,sáng mai đi học rùi.cảm ơn ạ
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Những câu hỏi liên quan
27 tháng 7 2016
\(\frac{2}{3}-\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{11}{2}-\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{2}{3}-\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{5}.\left(\frac{3.x}{2}-\frac{1}{4}\right)=\frac{-55}{12}\)
\(\Leftrightarrow\frac{3.x}{2}-\frac{1}{4}=\frac{-55}{12}:\frac{1}{5}\)
\(\Leftrightarrow\frac{3.x}{2}-\frac{1}{4}=\frac{-275}{12}\)
\(\Leftrightarrow\frac{3.x}{2}=\frac{-275}{12}+\frac{1}{4}\)
\(\Leftrightarrow\frac{3.x}{2}=\frac{-68}{3}\)
\(\Leftrightarrow\left(3.x\right).3=-136\)
\(\Leftrightarrow3.x=-136:3\)
\(\Leftrightarrow3.x=\frac{-136}{3}\)
\(\Leftrightarrow x=\frac{-136}{3}:3\)
\(\Leftrightarrow x=\frac{-136}{9}\)
12 tháng 8 2018
a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
T
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 4 2023
A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
NM
tính nhanh (2/3+3/4+5/6+...+99/100).(1/2+2/3+3/4+...+98/99)-(1/2+1/3+...+99/100).(2/3+2/4+...+98/99)
0
NX
16 tháng 8 2023
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
NX
16 tháng 8 2023
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
LN
1
R
5 tháng 2 2022
Ta có \(63,1.2-21,3.6=0,9.7.10.1,2-21.3,6\)
\(=6,3.1,2-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.1,2-6,3.1,2\)
\(=0\)
\(\Rightarrow\dfrac{\left(1+2+......+100\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}=\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+......+99-100}=0\)
Đặt A = \(\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
\(\Rightarrow3A=\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)
\(\Rightarrow3A+A=\left(\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow4A=\dfrac{1}{3}-\dfrac{1}{3^{101}}\Rightarrow A=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{101}}}{4}\)
Bạn tham khảo nhé.
Mình gọi A là tổng nhé.
Ta có:
A=13−132+133−134+...+1399−13100A=13−132+133−134+...+1399−13100
=> 3A=1−13+132−133+...+1398−13993A=1−13+132−133+...+1398−1399
=> A+3A=1−13100A+3A=1−13100
=> 4A=3100−131004A=3100−13100
=> A=3100−14.3100