Tính tổng :
a, 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192 + 1/ 384
b, 1/4x7 + 1/7x10 + 1/10x13 +...........+ 1/19x22
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1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192 + 1/384 = 85/128
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}+\frac{1}{768}+\frac{1}{1536}\)
\(A\times2=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}+\frac{2}{384}+\frac{2}{768}+\frac{2}{1536}\)
Rút gọn ta được
\(A\times2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}+\frac{1}{768}\)
\(A\times2-A=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{768}-\left[\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1536}\right]\)
\(A=\frac{2}{3}+\frac{1}{3}-\frac{1}{3}-\frac{1}{1536}\)
\(A=\frac{2}{3}-\frac{1}{1536}=\frac{341}{512}\)
1/1.4+1/4.7+1/7.10+1/10.13+1/13.16
=1/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16)
=1/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)
=1/3.(1/1-1/16)
=1/3.(16/16-1/16)=1/3.15/16=5/16
\(C=\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{2020+2023}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{2020.2023}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{3}.\dfrac{2019}{8092}\)
\(=\dfrac{673}{8092}\)
=1−14 +14 −110 +...+119 −122
=1−122
= \(\frac{21}{22}\)
k cho mình nha chắc chắn đúng 100 %
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\)
\(=1-\frac{1}{22}\)
\(=\frac{21}{22}\)
\(=2\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{192}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{192}\right)\)
\(=2\left(1-\frac{1}{192}\right)\)
\(=2\times\frac{191}{192}\)
\(=\frac{191}{96}\)
\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+....+\frac{1}{384}\)
\(\text{Đ}\text{ặt}\)\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{384}\)
\(2A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{192}\)
\(2A-A=\frac{1}{3}-\frac{1}{384}\)
a đề sai
b)Đặt A=1/4x7 + 1/7x10 + 1/10x13 +...........+ 1/19x22
\(3A=3\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{19.22}\right)\)
\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\)
\(3A=\frac{1}{4}-\frac{1}{22}\)
\(A=\frac{9}{44}:3\)
\(A=\frac{3}{44}\)