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Đặt A=1+4+42+.....+4200+4201
4A=4+42+43+.....+4201 +4202
4A-A= (4+42+43+.....+4201 +4202)-(1+4+42+.....+4200+4201)
3A= 4+42+43+.....+4201 +4202-1-4-42-.....-4200-4201
3A= 4202-1
A=\(\frac{4^{202}-1}{3}\)
Ta có:
B=1/2-1/2^2-1/2^3-...-1/2^100
B/2=1/2^2-1/2^3-1/2^4-....-1/2^101
B/2-B=1/2^101-1/2
=>B=(1/2^101-1/2).2
Vậy:B=(1/2^101-1/2).2
\(G=\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2005}}\)\(\Rightarrow3G=3+\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow3G-G=2G=3-\frac{1}{3^{2005}}\)\(\Rightarrow G=\frac{3-\frac{1}{3^{2005}}}{2}\)
\(Y=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)\(\Rightarrow2Y=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2Y-Y=2-\frac{1}{2^{2012}}\) \(\Rightarrow Y=2-\frac{1}{2^{2012}}\)
a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6D=1-\frac{1}{7^{100}}\)
\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)
B = 1 + 4 + 42 +...+ 4200 + 4201
=> 4B = 4 + 42 +43 +...+ 4201 + 4202
=> 4B-B = 4202 - 1
3B = 4202 -1
\(\Rightarrow B=\frac{4^{202}-1}{3}\)
4B = 4 + 4^2 + 4^3 + ... + 4^202
4B - B = ( 4 + 4^2 + 4^3 + ... + 4^202 ) - ( 1 + 4 + 4^2 + ... + 4^201 )
3B = 4^202 - 1
B = \(\frac{4^{202}-1}{3}\)
Tính nhanh:
\(A=\frac{2}{1+2}+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)
Đặt \(A=\frac{2}{1+2}+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)
\(=2-1+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)
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\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+....+\frac{2+3+...+20}{1+2+3+...+20}\)
\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)
\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)
\(A=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{210}\right)\)
\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)
\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)
\(A=19-\left[2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\right]\)
\(A=19-\left[2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\right]\)
\(A=19-\left[2\cdot\frac{19}{42}\right]=19-\frac{19}{21}=\frac{380}{21}\)
Vậy A = .....