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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
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
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<\(\frac{1}{2}\)
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
\(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{100-1}{100!}\)
\(=\frac{2}{2!}-\frac{1}{2!}+...+\frac{100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100!}\)
\(=1-\frac{1}{100!}=\frac{99!}{100!}< 1\)
\(A=\frac{1}{2\text{!}}+\frac{2}{3\text{!}}+...+\frac{99}{100\text{!}}=\frac{2-1}{2\text{!}}+\frac{3-1}{3\text{!}}+...+\frac{100-1}{100\text{!}}\)
\(=\frac{2}{2\text{!}}-\frac{1}{2\text{!}}+\frac{3}{3\text{!}}-\frac{1}{3\text{!}}+...+\frac{100}{100\text{!}}-\frac{1}{100\text{!}}\)
\(=1-\frac{1}{2\text{ }}+\frac{1}{2}-...-\frac{1}{100\text{!}}\)
\(=1-\frac{1}{100\text{!}}=\frac{99}{100\text{!}}< 1\)
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}< \frac{3}{4}\left(đpcm\right)\)