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sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
Đặt \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(\Rightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(\Rightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
\(\Rightarrow A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(\Rightarrow A=\frac{1}{1!}-\frac{1}{100!}\)
\(\Rightarrow A=1-\frac{1}{100!}\)
Mà \(1-\frac{1}{100!}< 1.\)
\(\Rightarrow A< 1\left(đpcm\right).\)
Chúc bạn học tốt!
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^2}+...+\frac{99}{3^{89}}-\frac{100}{3^{99}}\)
\(\Rightarrow4C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\left(1\right)\)
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 :))
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
\(\frac{1}{2!}+\frac{2}{3!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{100-1}{100!}\)
= \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{99!}-\frac{1}{100!}\)
= \(1-\frac{1}{100!}
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
= \(\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+....+\frac{100-1}{100!}\)
= \(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+....+\frac{100}{100!}-\frac{1}{100!}\)
= \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
= \(1-\frac{1}{100!}
Ta có : \(VT=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{100-1}{100!}\)
\(=\frac{2}{1.2}-\frac{1}{2!}+\frac{3}{1.2.3}-\frac{1}{3!}+\frac{4}{1.2.3.4}-\frac{1}{4!}+\frac{5}{1.2.3.4.5}-\frac{1}{5!}+...+\frac{100}{1.2...99.100}-\frac{1}{100!}\)
\(=\frac{1}{1}-\frac{1}{2!}+\frac{1}{1.2}-\frac{1}{3!}+\frac{1}{1.2.3}-\frac{1}{4!}+\frac{1}{1.2.3.4}-\frac{1}{5!}+...+\frac{1}{1.2...99}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)