B=1/1.3+1/3.5+...+1/2015.2017
Những câu hỏi liên quan
1/1.3+1/3.5+1/5.7+....+1/2015.2017
Dễ thôi:
Khoảng cách là 2
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\frac{1}{2}.\left(1-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
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cảm ơn bạn đã giúp mình!!
1/1.3+1/3.5+.....+1/2015.2017
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
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\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}\)
\(=\frac{1008}{2017}\)
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\(=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{2015.2017}\right)\)\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2015.2017}\right)\)\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{2015}-\frac{1}{2017}\right)\)\(=\frac{1}{2}\left(1-\frac{1}{2017}\right)\)\(=\frac{1}{2}.\frac{2016}{2017}\)\(=\frac{1008}{2017}\)
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Xem thêm câu trả lời
A= 1/2 ( 1 + 1.3) ( 1 + 1/2.4) ( 1 + 3.5 ) ...............( 1 + 1/ 2015.2017)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
mk đầu tiên đấy
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Xem thêm câu trả lời
-2/1.3-2/3.5-2/5.7-2/7.9-.....-2/2015.2017-1/27
B = (1+1/1.3)(1+1/2.4)(1+1/3.5)......(1+1/2014.2016)(1+ 1/2015.2017)
Nếu có thể thì các bạn giải ra giùm mik nha
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mấy chế ko giúp thì thôi lại còn nhấn linh tinh
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câu 1 tính
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2015.2017}\right)\)
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\) \(=\dfrac{2016}{2017}\)
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\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{3.5}\right).\left(1+\frac{1}{5.7}\right).....\left(1+\frac{1}{2015.2017}\right)\) 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
\(ChoA=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{3.5}\right).\left(1+\frac{1}{5.7}\right).....\left(1+\frac{1}{2015.2017}\right)\)
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