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=1/2 + (1/2*3+1/3*4) + (1/4*5+1/5*6) + (1/6*7+1/7*8) + (1/8*9+1/9*10)
=1/2 + 1/2*3.(1+1/2) + 1/2*5.(1/2+1/3) + 1/2*7.(1/3+1/4) + 1/2*9.(1/4+1/5)
=1/2 + 1/2*3.(3/2) + 1/2*5.(5/6) + 1/2*7.(7/12) + 1/2*9.(9/20)
=1/2 + 1/4 + 1/12 + 1/24 + 1/40
=9/10
6 = 2 x 2 + 2 = 2(2+1)
12 = 3 x 3 + 3 = 3(3+1)
20 = 4 x 4 + 4 = 4(4+1)
...
90 = 9 x 9 + 9 = 9(9+1)
Ta có đồng nhất thức sau
1/[n(n+1)] = 1/n - 1/(n+1)
Vậy
1/6 = 1/2 - 1/3
1/12 = 1/3 - 1/4
1/20 = 1/4 - 1/5
....
1/90 = 1/9 - 1/110
S = 1/2 -1/3+1/3-1/4+..............+1/9 -1/10
Tổng là 1/2 + 1/2 - 1/10 = 1 - 1/10 = 9/10
S= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
S= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 -1/6 +1/6 - 1/7 + 1/7 - 1/8
S= 1/2 - 1/ 8
S= 3/8
S= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
= 1/2 - 1/3 + 1/3 - ...+ 1/7 - 1/8
= 1/2 - 1/8
= 3/8
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{45.46}\)
\(\Rightarrow S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{45.46}\)
\(\Rightarrow S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{46}\)
\(\Rightarrow S=1-\frac{1}{46}\)
\(\Rightarrow S=\frac{45}{46}\)
Bài làm
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{45.46}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{45.46}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{46}\)
\(S=\frac{1}{1}-\frac{1}{46}\)
\(S=\frac{46}{46}-\frac{1}{46}\)
\(S=\frac{45}{46}\)
Vậy \(S=\frac{45}{46}\)
# Học tốt #
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{650}\) = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{25.26}\)= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}\)= \(1-\frac{1}{26}=\frac{25}{26}\)
~Học tốt~
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{650}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{25.26}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{25}-\frac{1}{26}\)
\(=1-\frac{1}{26}=\frac{25}{26}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
= \(\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)8 số hạng 1
= \(\left(1.8\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
= \(8-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(8-\left(1-\frac{1}{10}\right)\)
= \(8-\frac{9}{10}\)
= \(\frac{71}{10}\)
Lời giải:
$D=(1+1+...+1)-(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{650})$
$=25-(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{25.26})$
$=25-(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26})$
$=25-(1-\frac{1}{26})=24+\frac{1}{26}$
\(S=\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{600}+\dfrac{1}{650}\)
\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}\)
\(=\dfrac{1}{2}-\dfrac{1}{26}=\dfrac{13-1}{26}=\dfrac{12}{26}=\dfrac{6}{13}\)