Tính
\(B=\frac{-1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
K
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KT
19 tháng 2 2018
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
19 tháng 2 2018
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)
3 tháng 7 2017
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
LA
13 tháng 1 2018
\(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+....+\frac{3}{1+2+3+...+100}\)
\(B=3+3\left(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+..+100}\right)\)
Xét thừa số tổng quát: \(\frac{1}{1+2+3+...+n}=\frac{1}{\left[\left(n-1\right):1+1\right]:2.\left(n+1\right)}=\frac{1}{\frac{n\left(n+1\right)}{2}}\)
Ta có: \(B=3+3\left(\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+\frac{1}{\frac{4\left(4+1\right)}{2}}+...+\frac{1}{\frac{100\left(100+1\right)}{2}}\right)\)
\(B=3+3\left[2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\right]\)
\(B=3+6\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(B=3+6\left(\frac{1}{2}-\frac{1}{101}\right)\)
M
0
Bài giải
Ta có : \(B=-\frac{1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }B=-\frac{1}{3^0}-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(B=-1-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
Đặt \(C=\frac{1}{3^1}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }3C=1+\frac{1}{3^1}+...+\frac{1}{3^{99}}\)
\(\Rightarrow\text{ }3C-C=2C=1-\frac{1}{3^{100}}\)
\(C=\frac{1-\frac{1}{3^{100}}}{2}=\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
Thay vào biểu thức B ta được :
\(B=-1-\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=-\frac{3}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=\frac{\left(-3\right)^{101}}{2\cdot3^{100}}-\frac{1}{2\cdot3^{100}}=\frac{\left(-3\right)^{101}-1}{2\cdot3^{100}}\)
Bài giải
Ta có : \(B=-\frac{1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }B=-\frac{1}{3^0}-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(B=-1-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
Đặt \(C=\frac{1}{3^1}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\)
\(\Rightarrow\text{ }3C=1+\frac{1}{3^1}+...+\frac{1}{3^{99}}\)
\(\Rightarrow\text{ }3C-C=2C=1-\frac{1}{3^{100}}\)
\(C=\frac{1-\frac{1}{3^{100}}}{2}=\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
Thay vào biểu thức B ta được :
\(B=-1-\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=-\frac{3}{2}-\frac{1}{2\cdot3^{100}}\)
\(B=\frac{\left(-3\right)^{101}}{2\cdot3^{100}}-\frac{1}{2\cdot3^{100}}=\frac{\left(-3\right)^{101}-1}{2\cdot3^{100}}\)