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\(1-3+3^2-3^3+....-3^{2007}+3^{2008}\)
\(3S=3-3^2+3^3-3^4+...-3^{2008}+3^{2009}\)
\(4S=3^{2009}+1\)
\(\Rightarrow A=4S-1-3^{2009}\)
\(=\left(3^{2009}+1\right)-1-3^{2009}\)
\(=0\)
Bài 1 :
\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\cdot\frac{24}{50}=1\)
\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)
#Louis
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{12}{25}x=1\)
Đến đây dễ rồi :)))
Bn tự tính típ nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
\(A=\frac{2^{100}-1}{2^{100}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
hok tốt!!
đặt A = \(1+2+2^2+...+2^{2008}\)
\(2A=2+2^2+...+2^{2009}\)
\(2A-A=2^{2009}-1\)
\(A=2^{2009}-1\)
=> \(=\frac{2^{2009}-1}{1-2^{2009}}\)
đặt tử \(A=1+2+2^2+2^3+...+2^{20098}\)
\(2a=2+2^2+2^3+...+2^{2009}\)
\(A=2^{2009}-1\)
\(b=-1\)
Tính tổng S=\(\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Làm giúp mk bài này nha!Cảm ơn mn nhiều:3
Mk chỉ làm đc bài 2 thôi!
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(\Rightarrow2S-S=6-\frac{3}{2^9}\)
\(\Rightarrow S=6-\frac{3}{2^9}\)
Chúc bạn học tốt ( sai thì đừng ném đá ) !
Ta có :
A = \(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}\)< \(\frac{1}{1.1}+\frac{1}{1.2}+...+\frac{1}{49.50}\)
A < \(1-1+1-\frac{1}{2}+...+\frac{1}{49}-\frac{1}{50}\)
A < 1 - 1/50 = 49/50 < 2
Vậy A < 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
\(A=\frac{M}{N}=\frac{2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}}\)\(=2009\)
Đặt \(A=1+2+2^2+2^3+...+2^{2008}\)
\(2A=2.\left(1+2+2^2+2^3+...+2^{2008}\right)\)
\(2A=2+2^2+2^3+...+2^{2009}\)\(2A-A=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)\)
\(A=2^{2009}-1\)
\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)
\(S=\frac{2^{2009}-1}{-\left(-1+2^{2009}\right)}=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=-1\)