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Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vậy \(A=\frac{49}{50}\)
Chúc bạn học tốt ~
A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
A= \(1-\frac{1}{50}\)
A= \(\frac{49}{50}\)
Lời giải:
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2014-2012}{2012.2013.2014}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{2012.2013}-\frac{1}{2013.2014}$
$=\frac{1}{1.2}-\frac{1}{2013.2014}< \frac{1}{2}$
$\Rightarrow A< \frac{1}{2}:2$
Hay $A< \frac{1}{4}$
Bài này mình chắc 100%, 1 đúng nha vì ghi cực khổ lắm:
1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}...+\frac{50-49}{49.50}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}
\(A=2.3+3.4+4.5+...+49.50\)
\(3A=2.3.3+3.4.3+4.5.3+...+49.50.3\)
\(3A=2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+49.50.\left(51-48\right)\)
\(3A=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+49.50.51-48.49.50\)
\(3A=-1.2.3+49.50.51\)
\(3A=-6+48450\)
\(3A=48444\)
\(A=\frac{48444}{3}\)
\(A=16148\)
Chúc bạn học tốt ~
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{25.26.27}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{25.26.27}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{25.26}-\frac{1}{26.27}\)
\(2B=\frac{1}{1.2}-\frac{1}{26.27}\)
\(2B=\frac{1}{2}-\frac{1}{702}\)
\(2B=\frac{175}{351}\)
\(B=\frac{175}{251}:2\)
\(B=\frac{175}{502}\)
Chúc bạn học tốt ~
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{5}{6}\)
Bài 5:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}< 1\)
Vậy A<1.
Bài 4: Bn ghi nhầm đề rồi.
Đề đúng: \(A=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2011.2013}\)
\(\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\)
\(\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(\frac{1}{2}A=1-\frac{1}{2013}\)
\(A=2.\frac{2012}{2013}=\frac{4024}{2013}\)
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