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\(\frac{5}{2\cdot4}+\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+.....+\frac{5}{48\cdot60}\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{50}\right)\)
Tự tính nốt :p
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\(E=2\times4+4\times6+6\times8+...+98\times100\)
\(6\times E=2\times4\times6+4\times6\times\left(8-2\right)+6\times8\times\left(10-4\right)+...+98\times100\times\left(102-96\right)\)
\(=2\times4\times6+4\times6\times8-2\times4\times6+...+98\times100\times102-96\times98\times100\)
\(=98\times100\times102\)
\(\Rightarrow E=\frac{98\times100\times102}{6}=166600\)
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\(A=2\times4+4\times6+6\times8+...+98\times100\)
\(6\times A=2\times4\times6+4\times6\times\left(8-2\right)+6\times8\times\left(10-4\right)+...+98\times100\times\left(102-96\right)\)
\(=2\times4\times6+4\times6\times8-2\times4\times6+6\times8\times10-4\times6\times8+...+98\times100\times102-96\times98\times100\)
\(=98\times100\times102\)
\(\Leftrightarrow A=\frac{98\times100\times102}{6}=166600\)
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Câu hỏi tương tự nha
= 2 x ( 2 + 2 ) + 4 x ( 2 + 2 ) + 6 x ( 2 +2 ) +....+98 x ( 98 + 2 )
= 2 x 2 + 2 x 2 + 2 x 4 +4+......+98 x 98 = 2 x 98
= 2 x ( 2 + 4 + 6 +....+98 ) +( 2 x 2 + 4x4 + 6 x 6 +...+98 x 98 )
= 2 x 2450 + 40425 x 4
= 4900 + 161700 = 166600
Gọi biểu thức trên là A ta có:
Zô câu hỏi tương tự là cách giải
ĐS A = 49/200
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Đặt:A = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
=> A = 2.(\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2008.2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))
=> A = 2.\(\frac{502}{1005}\)
=> A = \(\frac{1004}{1005}\)
đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\)
=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
=\(\frac{1}{2}-\frac{1}{2010}\)
=\(\frac{502}{1005}\)
Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)
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a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)
Tổng A là: (2020+5)x404:2=409050
b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)
Vậy .....
A = 5 + 10 + 15 + ... + 2015 + 2020
Số số hạng là : 404
A = 409050
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
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S=(2+98)*(4+6)+...+100+100+102
100*10+....+100+100*102
=224400
Đặt \(A=\frac{5}{4.6}+\frac{5}{6.8}+.....+\frac{5}{48.50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+.....+\frac{2}{48.50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{48}-\frac{1}{50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{1}{4}-\frac{1}{50}\)
\(\Leftrightarrow\frac{5}{2}A=\frac{23}{100}\Rightarrow A=\frac{23}{100}.\frac{2}{5}=\frac{23}{250}\)