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Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\)
\(=\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+\left(\frac{1}{3}+\frac{1}{96}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)
\(=\frac{99}{1.98}+\frac{99}{2.97}+\frac{99}{3.96}+...+\frac{99}{49.50}\)
\(=99\left(\frac{1}{1.98}+\frac{1}{2.97}+\frac{1}{3.96}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4....98\)
\(=99\left(\frac{1}{1.98}+\frac{1}{2.97}+\frac{1}{3.96}+...+\frac{1}{49.50}\right).2.3.4....98\)chia hết cho 99 (đpcm)
\(=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+....+\left(\frac{1}{49}+\frac{1}{50}\right)=\frac{99}{1\times98}+\frac{99}{2\times97}+.....\frac{99}{49\times50}\)
Ta gọi các thừa số phụ là : \(a_1,a_2,......,a_{49}\)
\(A=\frac{99\times\left(a_1+a_2+.....+a_{49}\right)}{2\times3\times......\times97\times98}\times2\times3\times......\times97\times98\)
\(A=99\times\left(a_1+a_2+.....+a_{49}\right)\)
\(\Rightarrow A:99\)
\(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)(Có 98 phân số => có 49 cặp)
\(=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}=99.\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
=> \(A=\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).1.2.3...98.99\)
=> A : 99 = \(\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).1.2.3...98=2.3.4...97+1.3.4..96.98+...+1.2.3..48.51...98\)
kết quả là số tự nhiên
=> A chia hết cho 99
\(\frac{1}{2}+\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}=\left(\frac{1}{1}+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)( có 98 phân số => có 8 cặp )
\(=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}=99.\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A=\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).1.2.3....98.99\)
\(\)A chia hết cho 99.
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)