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\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
\(C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow3C=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
Trừ dưới cho trên:
\(2C=1+\frac{2}{3}-\frac{1}{3}+\frac{3}{3^2}-\frac{2}{3^2}+\frac{4}{3^3}-\frac{3}{3^3}+...+\frac{100}{3^{99}}-\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=B\Rightarrow2C=B-\frac{100}{3^{100}}\)
\(B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3B-3+\frac{1}{3^{99}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=B\)
\(\Rightarrow2B=3-\frac{1}{3^{99}}\Rightarrow B=\frac{3}{2}-\frac{1}{2.3^{99}}< \frac{3}{2}\)
\(\Rightarrow2C=B-\frac{100}{3^{100}}< B< \frac{3}{2}\Rightarrow C< \frac{3}{4}\)
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3A-A=\(1-\frac{1}{3^{99}}\)
2A=\(1-\frac{1}{3^{99}}\)
vì 2A<1
=> A<\(\frac{1}{2}\)
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^2}+...+\frac{99}{3^{89}}-\frac{100}{3^{99}}\)
\(\Rightarrow4C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\left(1\right)\)
Đặt: \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\Rightarrow3B=2+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow4C< B< \frac{3}{4}\)
\(\Rightarrow C< \frac{3}{16}\left(đpcm\right)\)
(Đánh nhanh quá sai chỗ nào thông cảm nha :))