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A = \(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{2013.2014.2015.2016}\)
3A =\(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{2013.2014.2015.2016}\)
3A = \(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{2013.2014.2015}-\dfrac{1}{2014.2015.2016}\)
3A = \(\dfrac{1}{1.2.3}-\dfrac{1}{2014.2015.2016}\)
3A = \(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}\)
A=\(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}:3\)
A=\(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}.\dfrac{1}{3}\)
A=\(\dfrac{2014.2015.2016-6}{9.2014.2015.2016}\)
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bn xem có giúp gì được ko nha!
http://toantieuhoclh.violet.vn/entry/show/entry_id/10236071
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Theo đầu bài ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow S=P\)
Vậy ( S - P )2016 = 02016 = 0
Mẫu số = 2015/1 + 2014/2 + 2013/3 + ... + 1/2015
= (1 + 1 + ... + 1) + 2014/2 + 2013/3 + ... + 1/2015
2015 số 1
= (2014/2 + 1) + (2013/3 + 1) + ... + (1/2015 + 1) + 1
= 2016/2 + 2016/3 + ... + 2016/2015 + 2016/2016
= 2016 × (1/2 + 1/3 + ... + 1/2015 + 1/2016)
=> phân số đề bài cho = 1/2016
1/2016