Cho \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{10000}\)
So sánh A với \(\frac{3}{4}\)
Những câu hỏi liên quan
Cho A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
So sánh A với 0,01
đặt C=1/2.3/4.5/6....9999/10000
đặt B=2/3.4/5.6/7....10000/10001
C.B=1/2.2/3.3/4.4/5.5/6.6/7....9999/10000.10000/10001
C.B=1/10001,đặt C.B=A
vì 1/10001<0,01=> A<0,01
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\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2007}{2^{2007}}\).
So sánh A với 2.
Tính nhanh
\(A=\frac{21}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}........\frac{9999}{10000}\)
\(C=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right).\left(1-\frac{1}{1000}\right)\)
CÁC BẠN GIẢI RA CHO MÌNH NHÉ
Tính giá trị của A biết:
1/ \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}\)
2/ \(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)
mọi nguời cứ tính đi đuợc câu nào tôi cũng tick cho
CMR:
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{25}+\frac{1}{36}+...+\frac{1}{10000}< \frac{25}{36}\)
Giải chi tiết giùm!
Cho \(A=\frac{3^3}{1}-\frac{5^3}{3}+\frac{7^3}{6}-\frac{9^3}{10}+\frac{11^3}{15}-\frac{13^3}{21}+\frac{15^3}{28}-\frac{17^3}{36}+...+\frac{199^3}{4950}\)
So sánh A với 814
Ta có:
\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+\frac{11^3}{30}-\frac{13^3}{42}+\frac{15^3}{56}-\frac{17^3}{72}+...+\frac{199^3}{9900}\)
\(=3^2.\left(1+\frac{1}{2}\right)-5^2.\left(\frac{1}{2}+\frac{1}{3}\right)+7^2.\left(\frac{1}{3}+\frac{1}{4}\right)-9^2.\left(\frac{1}{4}+\frac{1}{5}\right)+...+199^2.\left(\frac{1}{99}+\frac{1}{100}\right)\)
\(=3^2+\left(\frac{3^2}{2}-\frac{5^2}{2}\right)-\left(\frac{5^2}{3}-\frac{7^2}{3}\right)+\left(\frac{7^2}{4}-\frac{9^2}{4}\right)-\left(\frac{9^2}{5}-\frac{11^2}{5}\right)+...+\left(\frac{197^2}{99}-\frac{199^2}{99}\right)+\frac{199^2}{100}\)
\(=3^2-8+8-8+...+8+\frac{199^2}{100}=3^2+\frac{199^2}{100}< 3^2+\frac{199.200}{100}=9+398=407\)
\(\Rightarrow A< 407.2=814\)
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A=\(\frac{9}{8}-\frac{8}{9}+\frac{3}{25}+\frac{1}{4}-\frac{5}{16}+\frac{19}{25}-\frac{1}{9}+\frac{2}{25}-\frac{1}{81}\)
GIẢI GIÚP MÌNK NHÉ!
\(A=\frac{9}{8}-\frac{8}{9}+\frac{3}{25}+\frac{1}{4}-\frac{5}{16}+\frac{19}{25}-\frac{1}{9}+\frac{2}{25}-\frac{1}{81}\)
\(A=\left(\frac{9}{8}+\frac{1}{4}-\frac{5}{16}\right)-\left(\frac{8}{9}+\frac{1}{9}-\frac{1}{81}\right)+\left(\frac{3}{25}+\frac{19}{25}+\frac{2}{25}\right)\)
\(A=\frac{17}{16}-\frac{80}{81}+\frac{24}{25}\)
\(A=\frac{33529}{32400}\)
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Cho A=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4026}\) , B=\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}\). So sánh \(\frac{A}{B}\) với \(1\frac{2013}{2014}\)
Chứng minh rằng
\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}
