1/2+1/6+1/12+1/20+1/30+....+1/990
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\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{990}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{6}{25}\)
=1/4.5+1/5.6+1/6.8+1/7.8+....1/33.30
=1/4-1/5+1/5-1/6+1/6-1/8+1/7-1/8+...+1/30-1/33
=1/4-1/33
=29/132
A=1/20+1/30+1/42+...+1/9900
=1/(4*5)+1/5*6)+1/(6*7)+...+1/(99*100)
=1/4-1/5+1/5-1/6+1/6-1/7+...+1/99-1/100
=1/4-1/100
=6/25
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{50}+\dfrac{1}{990}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{50}+\dfrac{1}{990}???\)
Quy luật của vế sau "..." sai, bạn xem lại đề bài!
Nếu đúng đề thì sẽ như sau:
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)
Đề bài đúng là như vậy.
Giải:
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}\)
\(=\dfrac{25-1}{100}\)
\(=\dfrac{24}{100}\)
\(=\dfrac{6}{25}\)
A=1/4*5 + 1/5*6 + 1/6*7 +.....+1/99*100
A=1/4-1/5+1/5-1/6+1/6-1/7+...+1/99-1/100
A=1/4-1/100
A=25/100-1/100
A=6/25
A = 1/20 + 1/30 + 1/42 + 1/56 + .........+1/990
A = 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + ...........+ 1/99.100
A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ....... + 1/99 - 1/100
A = 1/4 - ( -1/5 + 1/5 ) - ( -1/6 + 1/6 ) - ( -1/7 + 1/7 ) - ...........- ( - 1/99 + 1/99 ) - 1/100
A = 1/4 - 0 - 0 - 0 - ........... - 0 - 1/100
A = 1/4 - 1/100
A = 25/100 - 1/100
A = 24/100
A = 6/25
Ta có\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{990}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{990}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)