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có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 + 2^10]
Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]
Q = 2 . 3+2^3 .3 +... + 2^9 .3
Q = 3. [ 2 + 2^3 +... + 2^9]
Vậy Q chia hết cho 3
Ta có: \(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Xét tử : \(2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+1+...+1\right)+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)( có 2008 số hạng 1 )
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)+1\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
Ghép tử và mẫu....
Vậy A = 2009
ta có: \(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+...+\frac{2007}{2^{2008}}\)
\(\Rightarrow S-\frac{1}{2}S=\frac{1}{2}+\left(\frac{2}{2^2}-\frac{1}{2^2}\right)+\left(\frac{3}{2^3}-\frac{2}{2^3}\right)+\left(\frac{4}{2^4}-\frac{3}{2^4}\right)+...+\left(\frac{2007}{2^{2007}}-\frac{2006}{2^{2007}}\right)-\frac{2007}{2^{2008}}\)
\(\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\)
Gọi \(Q=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2007}}\)
\(\Rightarrow\frac{1}{2}Q=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2008}}\)
\(\Rightarrow Q-\frac{1}{2}Q=\frac{1}{2}-\frac{1}{2^{2008}}\)
\(\Rightarrow\frac{1}{2}Q=\frac{1}{2}-\frac{1}{2^{2008}}\)
\(Q=\left(\frac{1}{2}-\frac{1}{2^{2008}}\right):\frac{1}{2}=1-\frac{1}{2^{2007}}\)
Thay Q vào S, ta có:
\(\frac{1}{2}S=1-\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\)
\(\Rightarrow S=\left(1-\frac{1}{2^{2007}}-\frac{2007}{2^{2008}}\right):\frac{1}{2}\)
\(S=2-\frac{1}{2^{2006}}-\frac{2007}{2^{2007}}< 2\)
\(\Rightarrow S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2007}{2^{2007}}< 2\)
tử là M mẫu là N ta dc
\(M=2008+\frac{2007}{2}+...+\frac{1}{2008}\)
\(=\left(1+...+1\right)+\frac{2007}{2}+...+\frac{1}{2008}\)
\(=\frac{2009}{2}+...+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)
vậy ta có
\(A=\frac{M}{N}=\frac{2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}}\)\(=2009\)