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$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)
=\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{1}{4}-\frac{1}{2.19.20}<\frac{1}{4}\)
B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}< 3\)
giong nhu dap an minh viet khi nay do
nho k cho minh voi nha
A=1/2(2/1.2.3+2/2.3.4+...+2/2014.2015.2016)~A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+...+1/2014.2015-1/2015.2016)~~A=1/2(1/1.2-1/2015.2016)~A=1/2(1/2-1/4062240)~A=1/2.2031119/4062240~A=203119/8124480. Dấu/= dấu gạch ps còn ~ là dấu xuống dòng. Còn bài này thì ko biết dung hay sai nua
Cho \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}.\)
Chứng minh rằng:\(B< \frac{1}{4}.\)
2B=\(\frac{2}{1.2.3}\)+.....+\(\frac{2}{18.19.20}\)
2B=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)+\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\).......+\(\frac{1}{18.19}\)-\(\frac{1}{19.20}\)
2B=\(\frac{1}{1.2}\)-\(\frac{1}{19.20}\)
B=\(\frac{1}{1.2}\):2-\(\frac{1}{19.20}\):2
B=\(\frac{1}{1.2}\).\(\frac{1}{2}\)-\(\frac{1}{19.20}\).\(\frac{1}{2}\)
=\(\frac{1}{4}\)-\(\frac{1}{19.20.2}\)<\(\frac{1}{4}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(2B=\frac{1}{1.2}-\frac{1}{19.20}\)
\(B=\left(\frac{1}{2}-\frac{1}{19.20}\right):2\)
\(B=\frac{189}{760}\)
2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 1/18.19.20
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 +...+1/18.19 - 1/19.20
2A = 1/1.2 - 1/19.20
2A = 1/2 - 1/19.20
A = (1/2 - 1/19.20) : 2
A = 1/4 - 1/(19.20.2)
MÀ 1/(19.20.2) > 0
nên A<1/4
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)
= 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)
= 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\)) - \(\frac{1}{21}\)
= 1 - \(\frac{1}{21}\)
= \(\frac{20}{21}\)< \(\frac{1}{4}\)
=> Đề bài có sai ko bạn?