B=1/1.3+1/3.5+...+1/2015.2017
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Dễ thôi:
Khoảng cách là 2
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\frac{1}{2}.\left(1-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}\)
\(=\frac{1008}{2017}\)
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
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9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\) \(=\dfrac{2016}{2017}\) 7 tháng 2 2020
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 |
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B=1/1.3+1/3.5+...+1/2015.2017
B= 1/2 . 2. ( 1/1.3+1/3.5 + .... + 1/2015 .2017)
B = 1/2 . ( 2/1.3 + 2/3.5 + ......+ 2/2015.2017)
B = 1/2. ( 1/1+ -1/3 + 1/3 + -1/5 + 1/5 +....+ 1/2015 + -1/2017)
B= 1/2 . ( 1/1 + -1 / 2017) = 1/2 . 2016 / 2017 = 2016 / 4034
Vậy B = 2016 / 4034 nha bn.pham hong thai