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\(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{42}\)+ .... + \(\frac{1}{132}\)
= \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ ... + \(\frac{1}{11.12}\)
= \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ ... + \(\frac{1}{11}\)- \(\frac{1}{12}\)
= \(\frac{1}{3}\)- \(\frac{1}{12}\)
= \(\frac{1}{4}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+.....+\frac{1}{132}\)
\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{11.12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}\)
\(=\frac{1}{4}\)
\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(B=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)
\(B=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(B=\frac{1}{4}-\frac{1}{12}\)
\(B=\frac{1}{6}\)
S=1/20+1/44+1/77+...+1/3080
S=1/20+1/44+1/77+...+1/3080
S.3/2=3/40+3/88+3/154+...+3/6160
S.3/2=3/40+3/88+3/154+...+3/6160
S.3/2=3/5.8+3/8.11+3/11.14+...+3/77.80
S.3/2=3/5.8+3/8.11+3/11.14+...+3/77.80
S.3/2=1/5−1/8+1/8−1/11+1/11−1/14+...+1/77−1/80
S.3/2=1/5-1/8+1/8-1/11+1/11-1/14+...+1/77-1/80
S.3/2=1/5−1/80
S.3/2=1/5-1/80
S.3/2=3/16
S.3/2=3/16
S=3/16:3/2
S=3/16:3/2
S=1/8
Tính tổng S=\(\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Làm giúp mk bài này nha!Cảm ơn mn nhiều:3
1-1/2+1/2-1/3+1/3+1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130
Ta có: \(B=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)
\(=\frac{4-3}{3\cdot4}+\frac{5-4}{4\cdot5}+...+\frac{11-10}{10\cdot11}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{11-3}{3\cdot11}=\frac{8}{33}\)
Vậy \(B=\frac{8}{33}\)
sai roi 0 co 1/30
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+.....+\frac{1}{132}\)
\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{11.12}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{3}-\frac{1}{12}\)
\(=\frac{1}{4}\)