\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
Giúp mk với...
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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
Chứng minh rằng: \(Q=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}<\frac{1}{2}\)
Mọi ngừi giúp mk với, mk cần gấp lắm hu hu
Nhân Q cho 3 ói lấy 3Q-Q sẽ ra 2Q=? =>Q òi so sánh
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Chứng tỏ giúp mình với !
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
Ta có \(A=\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{99}{100}}\)
\(A=\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{100}\right)}{\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)}\)
\(A=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\)
\(\Rightarrow A=2\)
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Ủa sao bạn ra được \(\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\) số 2 ở 200 đâu ra vậy ! và \(\frac{3}{2}\)nữa !
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Tính Q=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}{\frac{100-1}{1}+\frac{102-2}{2}+...+\frac{100-99}{99}}\)
Sửa đề:
\(Q=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{100-1+\frac{100}{2}-1+...+\frac{100}{99}-1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{100}{100}+\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{100.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{100}\)
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Tính:
\(\left[\frac{1}{100}-1^2\right].\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right].\left[\frac{1}{100}-\left(\frac{1}{3}\right)^2\right].....\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)
Giải nhanh lên giúp mk với! Rồi mk tick cho 3 cái
đây có chắc là toán lớp 7 không đấy
nếu có bài hình nào khó thì cho lên đấy nhé mình chuyên về toán lớp 7 hơn
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GửiTNs tao cuồng:c/m Bfrac{1}{2}+frac{1}{2^2}+frac{3}{2^3}+....+frac{100}{2^{100}}2Ta có:2B1+frac{1}{2}+frac{3}{2^2}+....+frac{100}{2^{99}}Rightarrow2B-BB1+left(frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+....+frac{1}{2^{99}}right)-frac{100}{2^{100}}(*)c/m frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+...+frac{1}{2^{99}}1Đặt Afrac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+....+frac{1}{2^{99}}Rightarrow2A1+frac{1}{2}+frac{1}{2^2}+...+frac{1}{2^{98}}Rightarrow2A-Aleft(1+frac{1}{2}+frac{1}{2^2}+...+frac{1}{2^{98}}right)-lef...
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Gửi
TNs tao cuồng:c/m \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{3}{2^3}+....+\frac{100}{2^{100}}<2\)Ta có:\(2B=1+\frac{1}{2}+\frac{3}{2^2}+....+\frac{100}{2^{99}}\)\(\Rightarrow2B-B=B=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)(*)c/m \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}<1\)Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)\)\(\Rightarrow A=1-\frac{1}{2^{99}}<1\)do đó \(B=1+A-\frac{100}{2^{100}}\Rightarrow B<2-\frac{100}{2^{100}}<2\left(đpcm\right)\)
Giúp vs: tính \(J=\frac{\frac{1}{99}+\frac{2}{98}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
Chứng minh rằng:a. frac{1}{3^2}+frac{2}{3^3}+frac{3}{3^4}+frac{4}{3^5}+...+frac{99}{3^{100}}+frac{100}{3^{101}} frac{1}{4}b.frac{1}{2}-frac{1}{4}+frac{1}{8}-frac{1}{16}+frac{1}{32}-frac{1}{64} frac{1}{3}c.frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{1}{16}d. frac{1}{5^2}-frac{2}{5^3}+frac{3}{5^4}-frac{4}{5^5}+...+frac{99}{5^{100}}-frac{100}{5^{101}} frac{1}{36}
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Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
\(\frac{\frac{1}{99}+\frac{2}{98}+....+\frac{98}{2}+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+...+\frac{1}{500}}\)