ta có:1/n(1+2+...+n)=1/n.n((n+1))/2=(n+1)/2

=>S=1+3/2+2+5/2+...+10=43

Bài 1: 

a: x+1/2=5/6

nên x=5/6-1/2=1/3

b: x+1/4=3/4

nên x=3/4-1/4=2/4=1/2

c: x+3/10=1/2

nên x=1/2-3/10=5/10-3/10=1/5

d: x+1/4=3/8

nên x=3/8-1/4=3/8-2/8=1/8

TOT LAM

M=\(\frac{1}{1\cdot2}\)- \(\frac{1}{2\cdot3}\)+ \(\frac{1}{2\cdot3}\) - \(\frac{1}{3\cdot4}\) +..........+ \(\frac{1}{10\cdot11}\) - \(\frac{1}{11\cdot12}\)

= \(\frac{1}{2}\) - \(\frac{1}{11\cdot12}\) 

=\(\frac{65}{132}\) 

a)  \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{19}{20}=\frac{1}{20}\)

b)  \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)

=>  \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

=> \(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

=>  \(A=2-\frac{1}{2^{2012}}\)

c) \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{231}{4}.\frac{4}{21}=11\)

d.e)  ktra lại đề

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)

\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+...+\frac{17}{2}=\frac{1}{2}\left(2+3+...+17\right)=\frac{1}{2}\cdot\frac{16.19}{2}=4.19=76\)

\(A=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{4950}=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)

\(=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+\frac{1}{\frac{4.\left(4+1\right)}{2}}+.....+\frac{1}{\frac{50\left(50+1\right)}{2}}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)