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Ta đã biết: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
Ta có: \(A=1+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+...+\frac{1}{20}.\left(\frac{20.21}{2}\right)\)
\(A=1+\frac{3}{2}+\frac{4}{2}+....+\frac{21}{2}\)
\(A=\frac{1}{2}.\left(2+3+....+21\right)\)
Tổng trong ngoặc có:21-2+2=20 (số hạng)
\(=>A=\frac{1}{2}.\left(\frac{\left(21+2\right).20}{2}\right)=\frac{1}{2}.230=115\)
Vậy..........
\(\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\left(-\frac{4}{3}\right)\)
\(=\left[\frac{2}{3}+1+1\right]:\left(-\frac{4}{3}\right)\)
\(=\frac{8}{3}.\frac{-3}{4}\)
\(=-2\)
help me T×m mét sè cã ba ch÷ sè, biÕt r»ng sè ®ã chia hÕt cho 18 vµ c¸c ch÷ sè cña nã tØ lÖ víi ba sè 1, 2 vµ 3.
Ta có, với \(n\) nguyên dương: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
Suy ra, \(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{1}{1+2}=\frac{1.4}{2.3}\)
\(1-\frac{1}{1+2+3}=\frac{2.5}{3.4}\)
....
\(1-\frac{1}{1+2+...+2013}=\frac{2012.2015}{2013.2014}\)
\(1-\frac{1}{1+2+...+2014}=\frac{2013.2016}{2014.2015}\)
Suy ra, \(P=\frac{\left(1.2.....2013\right).\left(4.5.....2016\right)}{2.\left(3.4.....2014\right)^2.2015}=\frac{2016}{3.2014}=\frac{336}{1007}\)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(A=\left(1-\frac{1}{\frac{\left(1+2\right).2}{2}}\right)\left(1-\frac{1}{\frac{\left(1+3\right).3}{2}}\right)...\left(1-\frac{1}{\frac{\left(1+2006\right).2006}{2}}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{2007.2006-2}{2006.2007}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}....\frac{2007.2006-2}{2006.2007}\) (1)
xét thấy:2007.2006-2=2006.(2008-1)+2006-2008=2006.(2008-1+1)-2008=2008.(2006-1)=2008.2005 (2)
(1),(2)\(=>A=\frac{4.1}{2.3}.\frac{5.2}{3.4}.\frac{6.3}{4.5}....\frac{2008.2005}{2006.2007}\)
\(A=\frac{\left(4.5.6...2008\right)\left(1.2.3...2005\right)}{\left(2.3.4....2006\right)\left(3.4.5...2007\right)}=\frac{2008}{2006.3}=\frac{1004}{3009}\)
Vậy A=1004/3009
bit lm bài này k giup tui
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)