Tính nhanh:
1.3+3.5+5.7 +......+2015.2017
1.3+3.5+5.7+..........+2015.2017
2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017
=1-1/3+1/3-1/5+...+1/2015-1/2017
=1-1/2017
=2016/2017
1/1.3+1/3.5+1/5.7+....+1/2015.2017
Dễ thôi:
Khoảng cách là 2
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\frac{1}{2}.\left(1-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
cảm ơn bạn đã giúp mình!!
2016/1.3 + 2016/3.5 + 2016/5.7 +...+ 2016/2015.2017=?
\(\frac{2016}{1.3}+\frac{2016}{3.5}+\frac{2016}{5.7}+....+\frac{2016}{2015.2017}\)
\(=1008.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=1008.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008.\left(1-\frac{1}{2017}\right)\)
\(=1008.\frac{2016}{2017}\)
-2/1.3-2/3.5-2/5.7-2/7.9-.....-2/2015.2017-1/27
Tính tổng :
M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.......+\frac{4}{2015.2017}\)
M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017
M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017
M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )
M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )
M = 2 . ( 1 - 1/2017 )
M = 2 . 2016/2017
M = 4032/2017
\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(M=2\left(1-\frac{1}{2017}\right)\)
\(M=\frac{4032}{2017}\)
\(M=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)
\(M=4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\right)\)
\(M=4.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015.2017}\right)\)
\(M=2\left(1-\frac{1}{2017}\right)\)
\(M=2.\frac{2016}{2017}\)
\(M=\frac{4032}{2017}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
mk đầu tiên đấy
A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+.....+\frac{3}{2015.2017}\)
cái đề này là tính nha mấy bạn
A=3/1*3+3/3*5+3/5*7+...+3/2015*2017
A=3/2*(2/1*3+2/3*5+2/5*7+...+2/2015*2017)
A=3/2*(1-1/3+1/3-1/5+1/5-1/7+...+1/2015-1/2017)
A=3/2*(1-1/2017)
A=3/2*2016/2017
A=3024/2017
A= \(\frac{3}{1.3}\)+\(\frac{3}{3.5}\)+\(\frac{3}{5.7}\)+....+\(\frac{3}{2015.2017}\)
A= \(\frac{3}{2}\).(\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+...+\(\frac{2}{2015.2017}\))
A= \(\frac{3}{2}\).( 1- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+... \(\frac{1}{2015}\)- \(\frac{1}{2017}\))
A= \(\frac{3}{2}\).(1- \(\frac{1}{2017}\))
A= \(\frac{3}{2}\). \(\frac{2016}{2017}\)
A= \(\frac{3024}{2017}\)
\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{3.5}\right).\left(1+\frac{1}{5.7}\right).....\left(1+\frac{1}{2015.2017}\right)\)
Tính A
Lời giải:
Xét tổng quát:
1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)
Thay k=1,2,....,2015k=1,2,....,2015 ta có:
1+11.3=221.31+11.3=221.3
1+12.4=322.41+12.4=322.4
1+13.5=423.51+13.5=423.5
1+14.6=524.61+14.6=524.6
.............
1+12015.2017=201622015.20171+12015.2017=201622015.2017
Nhân theo vế:
⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)
=12.221.3.322.4.423.5.524.6....201622015.2017=12.221.3.322.4.423.5.524.6....201622015.2017
=(1.2.3...2016)2(1.2.3...2015)(2.3.4...2017)=(1.2.3...2016)(2.3....2016)(1.2.3...2015)(2.3.4...2017)=2016.12017=20162017