Giải dùm mình
1 trên 3 nhâb 5 cộng 1 trên 5 nhân 7 cộng ... cộng 1 trên 97 nhân 99
B = -1 phần 3 cộng 1 phần 3 mũ 2 -1 phần 3 mũ 3 cộng ... cộng 1 phần 3 mũ 50 trừ 1 phần 3 mũ 51
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\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{1}{2}-\frac{1}{2.3^{100}}\)
Đặt A= \(\frac{1}{2}\)-\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)-\(\frac{1}{2^2}\)+....+\(\frac{1}{2^2}\)
=> 2A=1-\(\frac{1}{2}\)+\(\frac{1}{2^2}\)-\(\frac{1}{23}\)+...+\(\frac{1}{2^{98}}\)
=> 2A+A=1+\(\frac{1}{2^{99}}\)
=> 3A=1+\(\frac{1}{2^{99}}\)
=> A= \(\frac{1}{3}\)+\(\frac{1}{3.2^{99}}\)
Đặt A =\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
Ta có \(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
=> \(2A=3A-A=3-\frac{1}{3^{2005}}\)
=> \(A-\frac{3-\frac{1}{3^{2005}}}{2}\)
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{2^3}<\frac{1}{2.3}\)
\(\frac{1}{2^4}<\frac{1}{3.4}\)
..........
\(\frac{1}{2^n}<\frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^n}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}\)
Mà \(1-\frac{1}{n}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+.....+\frac{1}{2^n}<1\left(đpcm\right)\)
a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
dễ, nhưng phai giai dc câu nay 60% nhan x cong 2 phan 3 = 1 phan 3 nhan 6va 1 phan 3
Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)
=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)