Tính:1/3+1/3^2+1/3^3+....+1/3^99
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1 tháng 5 2017
\(G=\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2005}}\)\(\Rightarrow3G=3+\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow3G-G=2G=3-\frac{1}{3^{2005}}\)\(\Rightarrow G=\frac{3-\frac{1}{3^{2005}}}{2}\)
\(Y=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)\(\Rightarrow2Y=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2Y-Y=2-\frac{1}{2^{2012}}\) \(\Rightarrow Y=2-\frac{1}{2^{2012}}\)
5 tháng 6 2018
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
5 tháng 6 2018
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
VT
1
3 tháng 10 2016
Tham khảo tại link này nha : 
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[ Toán 7] Tính A: $A=1+3+3^2+....+3^{100} $ | HOCMAI Forum - Cộng đồng học sinh Việt Nam
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Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!
H9
HT.Phong (9A5)
CTVHS
26 tháng 3 2023
\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)
\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)
\(=0\)
HT
3
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
24 tháng 6 2023
1/2* x+2/3=9/2
1/2 * x = 9/2 - 2/3
1/2 * x= 23/6
x= 23/6 : 1/2
x= 23/6 x 2= 23/3
___
1/2*x-1/3=2/3
1/2*x = 2/3 + 1/3
1/2 * x= 1
x= 1: 1/2
x= 2
____
1/4+3/4:x=3
3/4 : x = 3 - 1/4
3/4 : x= 11/4
x= 11/4 : 3/4
x= 11/3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 6 2023
\(\dfrac{1}{2}\)\(\times\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{9}{2}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) = \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) = \(\dfrac{23}{6}\)
\(x\) = \(\dfrac{23}{6}\):\(\dfrac{1}{2}\)
\(x\) = \(\dfrac{23}{3}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) = \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\times\)\(x\) = 1
\(x\) = 1 : \(\dfrac{1}{2}\)
\(x\) = 2
\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\): \(x\) = 3
\(\dfrac{3}{4}\): \(x\) = 3 - \(\dfrac{1}{4}\)
\(\dfrac{3}{4}\):\(x\) = \(\dfrac{11}{4}\)
\(x\) = \(\dfrac{3}{4}\): \(\dfrac{11}{4}\)
\(x\) = \(\dfrac{3}{11}\)
AN
1
26 tháng 3 2022
a)
(1/3-1/5)x1/4 = 1/3x1/4-1/5x1/4=1/12-1/20=1/30
(1/3-1/5)x1/4 = 2/15x1/4=1/30
b)
2/5 x 3/7 + 2/5 x 4/7= 6/35 + 8/35 = 14/35 = 2/5
2/5 x 3/7 + 2/5 x 4/7= (3/7+4/7) x 2/5 = 1 x 2/5 = 2/5
\(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow2S=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2S=1-\frac{1}{3^{99}}\)
\(\Rightarrow S=\frac{1-\frac{1}{3^{99}}}{2}\)
Đặt \(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{99}}\)
\(\Rightarrow3S=1+\frac{1}{3}+\frac{1}{3^2}+.......+\frac{1}{3^{98}}\)
\(\Rightarrow3S-S=\left(1+\frac{1}{3}+.....+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2S=1-\frac{1}{3^{99}}\Rightarrow S=\frac{1-\frac{1}{3^{99}}}{2}=\frac{\frac{3^{99}-1}{3^{99}}}{2}=\frac{3^{99}-1}{3^{99}.2}\)