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\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
\(M=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(\Rightarrow M=\frac{1}{2}.\frac{32}{99}\)
\(\Rightarrow M=\frac{16}{99}\)
\(M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(M=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5.}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{16}{99}\)
Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99
=2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
=1+0+0+0+...+0+1/99
=1+1/99
=100/99
Mà 100/99>1.Suy ra S>1
Vậy S>1
M=(1/3-1/5)+(1/5+1/7)+...+(1/97+1/99)
M=1/3+(1/5-1/5)+...+(1/97-1/97)-1/99
M=1/3-1/99
M=32/99
Đây là bài toán tìm tổng dãy số có quy luật.
Để ý thấy rằng \(\frac{1}{n\left(n+2\right)}=\frac{1}{2}.\frac{2}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)\)
Vậy thì \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{n+2}\right)=\frac{5}{36}\Rightarrow\frac{1}{3}-\frac{1}{n+2}=\frac{5}{18}\)
\(\Rightarrow\frac{1}{n+2}=\frac{1}{18}\Rightarrow n=16.\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{n\left(n+2\right)}=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{1}{3}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{12}{36}-\frac{1}{n+2}=\frac{5}{36}\)
\(\frac{1}{n+2}=\frac{7}{36}\)
\(\Rightarrow\frac{7}{7\left(n+2\right)}=\frac{7}{36}\)
\(7\left(n+2\right)=36\)
n + 2 = 36/7
n = 36/7 - 2
( Tự tính KQ nha )
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1 /7 +.... + 1/97 - 1/99
=1/3 - 1/99
=32/99
m=/3.5+2/5.7+2/7.9+.....+2/97.99
=m=1/3-1/5+1/5-1/7+.......+1/97-1/99
m=1/3-1/99
=32/99