Tính :
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(C=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
Những câu hỏi liên quan
1/ Cho A dfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+.....+dfrac{99}{3^{99}}-dfrac{100}{3^{100}} Chứng minh A dfrac{3}{16}2/ Cho B(dfrac{1}{2^2}-1)(dfrac{1}{3^2}-1)....(dfrac{1}{100^2}-1) So sánh B và dfrac{-1}{2}
Đọc tiếp
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
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Tính
\(C=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+..+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
9 tháng 5 2021 lúc 20:08
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
Mình làm được một câu thôi, bạn dựa vào làm nha!
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6 tháng 8 2021 lúc 17:11
a) CMR: \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
b) Tính: \(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)
Nếu \(a+b+c=0\)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+0.\dfrac{2}{abc}\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
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a. Đề bài sai, chắc chắn thiếu dữ kiện \(a+b+c=0\)
b.
\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{\left(-100\right)^2}}=\sqrt{\left(1+\dfrac{1}{99}-\dfrac{1}{100}\right)^2}\)
\(=1+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{9901}{9900}\)
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Lời giải:
a.
Cho $a=1; b=5; c=6$ thì thấy đề sai.
b.
\(1+\frac{1}{99^2}+\frac{1}{100^2}=(1+\frac{1}{99})^2-\frac{2}{99}+\frac{1}{100^2}\)
\(=(\frac{100}{99})^2-\frac{2}{99}+\frac{1}{100^2}=(\frac{100}{99}-\frac{1}{100})^2\)
\(\Rightarrow \sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=\frac{100}{99}-\frac{1}{100}\)
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B(1-dfrac{1}{2^2})(1-dfrac{1}{3^2})(1-dfrac{1}{4^2})(1-dfrac{1}{5^2})...(1-dfrac{1}{99^2})(1-dfrac{1}{100^2})C(dfrac{1}{4}-1)(dfrac{1}{9}-1)(dfrac{1}{16}-1)...(dfrac{1}{999}-1)
Đọc tiếp
B=(1-\(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))(1-\(\dfrac{1}{5^2}\))...(1-\(\dfrac{1}{99^2}\))(1-\(\dfrac{1}{100^2}\))
C=(\(\dfrac{1}{4}\)-1)(\(\dfrac{1}{9}\)-1)(\(\dfrac{1}{16}\)-1)...(\(\dfrac{1}{999}\)-1)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
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tính H = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
25 tháng 7 2021 lúc 11:49
Chứng minh : \(\dfrac{99}{100}\) > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},...,\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(A\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A<\(1-\dfrac{1}{100}=\dfrac{99}{100}\)(đpcm)
Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2.3},\dfrac{1}{3^2}>\dfrac{1}{3.4},...,\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
A>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
A>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
A>\(\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)(đpcm)
Vậy \(\dfrac{99}{100}>A>\dfrac{99}{202}\)
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Tính giá trị biểu thức \(P=\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\).
\(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{\left(n+1\right)^2}}\\ =\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+\dfrac{2}{n}-\dfrac{2}{n+1}-\dfrac{2}{n\left(n+1\right)}}\\ =\sqrt{\left[1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right]^2}=\left|1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right|\)
\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)
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