Tính \(M=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
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VD
1
T
17 tháng 2 2017
ĐẶT A=\(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\frac{1}{3}A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\frac{1}{3}A-A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)
\(\frac{-2}{3}A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)
\(A=\frac{\frac{1}{3^{2006}}-1}{\frac{-2}{3}}\)
13 tháng 3 2020
\(A=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow3A=1+\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)
\(\Rightarrow A=\frac{3^{2005}-1}{3^{2005}.2}\)
6 tháng 8 2016
Đặt A \(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2005}}\right):2\)
11 tháng 7 2018
a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
XO
11 tháng 10 2020
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
11 tháng 10 2020
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
TP
12 tháng 7 2017
Ta có:
\(\frac{1\div2003+1\div2004-1\div2005}{5\div2003+5\div2004-5\div2005}\) - \(\frac{2\div2002+2\div2003-2\div2004}{3\div2002+3\div2003-3\div2004}\)
Đơn giản đi hết ta sẽ còn:
\(\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
2.
Ta có:
Số khoảng cách của các số trong dãy là 23 = 8
=> Tổng của dãy dưới sẽ gấp 8 lần tổng dãy trên.
=> 3025 . 8 = 24200
23 tháng 12 2016
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
T
23 tháng 12 2016
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
LT
2
17 tháng 6 2018
\(3S=3+\frac{1}{3}+...+\frac{1}{3^{2004}}\)
\(3S-S=\left(3+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(2S=3-\frac{1}{3^{2005}}\)
\(2S=\frac{3^{2006-1}}{3^{2005}}\)
\(S=\frac{3^{2006}-1}{3^{2005}.2}\)
17 tháng 6 2018
S = 1/3 + 1/32 + 1/33 + ... + 1/32005
=> 3S = 1 + 1/3 + 1/32 + ... + 1/32004
=> 3S - S = 1 + 1/3 + 1/32 + ... + 1/32004 - (1/3 + 1/32 + 1/33 + ... + 1/32005)
=> 2S = 1 + 1/3 + 1/32 + ... + 1/32004 - 1/3 - 1/32 - 1/33 - ... - 1/32005
=> 2S = 1 - 1/32005
=> S = \(\frac{\frac{1}{3^{2005}}}{2}\)
=> S = 1/32005.2