2/1*3+2/3*5+2/5*7+...+2/99*101 = ?
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2/1×3+2/3×5+.......+2/99×101
5/1×3+5/3×5+5/5×7+.......+5/99×101
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)
\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)
\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)
....
\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
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\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)
=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)
=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)
=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)
=\(\frac{5}{2}\cdot\frac{100}{101}\)
\(=\frac{250}{101}\)
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= 3 - 1 / 1 x 3 + 5 - 3 / 3 x 5 + ... + 101 - 99 / 99 x 101
= 1 - 1 / 3 + 1 / 3 - 1 / 5 + 1 / 5 - ... - 1 / 99 + 1 / 99 - 1 / 101
gạch gạch gạch gạch ... gạch gạch
= 1 - 1 / 101
= 100 / 101
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Bài tập *
a) 2/1×3 + 2/3×5 + 2/5×7 + ...............+ 2/99×101
b) 5/1×3 + 5/3×5 + 5/ 5×7 + ................ + 5/99×101
Giúp mk bài này nhé mk
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
2/1×3+2/3×5+2/5×7+2/7×+11+........+2/99×101
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
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Tính B=1*3+5*7+9*11+...+97*101
C=1*3*5-3*5*7+5*7*9-....-97*99*101
D=1*99+3*97+5*95+...+49*51
E=1*3^3+3*5^3+5*7^3+...+49*51^3
F=1*99^2+2*98^2+3*97^2+...+49*51^2
cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà
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2/1×3+2/3×5+2/5×7+...+2/99×101
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
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2/1×3+2/3×5+2/5×7+...+2/99×101
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
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Ta có : \(\frac{2}{1.3}\)= \(\frac{3-1}{1.3}\)=\(\frac{3}{1.3}\)- \(\frac{1}{1.3}\)= 1 - \(\frac{1}{3}\)
\(\frac{2}{3.5}\) = \(\frac{5-3}{3.5}\)=\(\frac{5}{3.5}\)- \(\frac{3}{3.5}\)=\(\frac{1}{3}\)- \(\frac{1}{5}\)
\(\frac{2}{5.7}\) = \(\frac{7-5}{5.7}\)= \(\frac{7}{5.7}\)- \(\frac{5}{5.7}\)= \(\frac{1}{5}\)- \(\frac{1}{7}\)
.........
\(\frac{2}{99.101}\) = \(\frac{101-99}{99.101}\)= \(\frac{101}{99.101}\)- \(\frac{99}{99:101}\)= \(\frac{1}{99}\)- \(\frac{1}{101}\)
= 1 - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+........ + \(\frac{1}{99}\)- \(\frac{1}{101}\)
= 1- \(\frac{1}{101}\)
= \(\frac{100}{101}\)
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2/1×3+2/3×5+2/5×7+...+2/99×101
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
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2/1×3 + 2/3×5 + 2/5×7+...+ 2/99×101
![꧁༺༒༻꧂[A]йɦ•βê•ʠǔá♡ (Hội...](https://hoc24.vn/images/avt/avt2989494_256by256.jpg)
\(\frac{2}{1}\)\(\times\)3+\(\frac{2}{3}\)\(\times\)5+\(\frac{2}{5}\)\(\times\)7+...+\(\frac{2}{99}\)\(\times\)101
=(1-\(\frac{1}{3}\))+(\(\frac{1}{3}\)-\(\frac{1}{5}\))+(\(\frac{1}{5}\)-\(\frac{1}{7}\))+...+(\(\frac{1}{99}\)-\(\frac{1}{100}\))
=1-\(\frac{1}{100}\)
=\(\frac{100}{101}\)
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TÍCH CHO MK NHA
CHÚC BẠN HỌC TỐT !
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