\(B=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)
\(B=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(B=2.\left(1-\dfrac{1}{10}\right)\)
\(B=2.\dfrac{9}{10}\)
\(B=\dfrac{9}{5}\)
anh ơi , đại học rồi mà ko giải đc bài này ạ?
B=1/1.2-1/2.3+1/2.3-1/3.4+...+1/8.9+1/9.10
B=1/1.2+0+0+...+0+1/9.10
B=1/1.2-1/9.10
=1/2-1/90
=44/90=22/45
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\)
⇔ \(2B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\)
\(2B=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+...+\dfrac{10-8}{8.9.10}\)
\(2B=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}+...+\dfrac{10}{8.9.10}-\dfrac{8}{8.9.10}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(2B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(2B=\dfrac{45}{90}-\dfrac{1}{90}\)
\(2B=\dfrac{44}{90}\)
\(B=\dfrac{44}{90}:2\)
\(B=\dfrac{44}{90}\times\dfrac{1}{2}\)
\(B=\dfrac{44}{180}=\dfrac{11}{45}\)
# Hạnh nè
Bài này em giải theo số hạng tổng quát em nhé
Mình xét \(\dfrac{2}{k\left(k+1\right)\left(k+2\right)}=\dfrac{k+2-k}{k\left(k+1\right)\left(k+2\right)}=\dfrac{1}{k\left(k+1\right)}-\dfrac{1}{\left(k+1\right)\left(K+2\right)}\) Do đó ta có
2B= \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+......+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)\(=\dfrac{1}{1.2}-\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{90}=\dfrac{43}{90}\) Từ đó suy ra B=43/180
