Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
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a) Cho A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}\)Chứng minh A<\(\frac{3}{4}\).
b) Chứng minh rằng:A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{99}}< \frac{1}{2}\)
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<\(\frac{1}{2}\)
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Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{99}{100}\)
Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)
=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)
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\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)
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chứng minh rằng:a/ \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\) \(\frac{1}{2}\)
b/\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)\(\frac{1}{2}\)
nhanh thì tích
chậm thì thôi
1.chứng minh rằng : \(\frac{1}{2}!+\frac{2}{3}!+\frac{3}{4}!+...+\frac{99}{100}!< 1\)
2. Chứng minh rằng :\(\frac{1.2-1}{2}+\frac{2.3-1}{3}+\frac{3.4-1}{4}+...+\frac{99.100-1}{100}< 2\)
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
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
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khi cộng cac số có tử bé hơn mẫu thì tổng sẽ <1 nha
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Chứng minh
100-(1-\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\))= \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Leftrightarrow99-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{100}\)
\(\Leftrightarrow1+1+1+...+1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)
\(\Leftrightarrow\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)\(\left(đpcm\right)\)
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Cho M =\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\) .Hãy chứng minh M<\(\frac{3}{16}\)
Câu 2 Chứng minh rằng :
\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
Tham khảo nha bạn :
Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến
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1. Chứng Minh Rằng \(\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}<\frac{3}{4}\)
2. Chứng Minh Rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2012}\)
2.
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\)
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Xem thêm câu trả lời
chứng minh rằng:
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=\frac{1}{2}+\frac{1}{3}+\frac{3}{4}+....+\frac{99}{100}\)
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\left(1+1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\) ( ĐPCM )
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Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
Giả sử \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(\Rightarrow100=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}+1+\frac{1}{2}+...+\frac{1}{100}\)
\(\Rightarrow100=1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)+...+\left(\frac{99}{100}+\frac{1}{100}\right)\)
\(\Rightarrow100=1+1+1+...+1\) (100 chữ số 1)
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
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Chứng minh rằng:
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
giúp minh với