Cho tổng T = \(\dfrac{2}{2^1}\)+\(\dfrac{3}{2^2}\)+\(\dfrac{4}{2^3}\)+..........+\(\dfrac{2016}{2^{2015}}\)+\(\dfrac{2017}{2^{2016}}\) So sánh T với 3
Những câu hỏi liên quan
cho tổng T= \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}\) +...+\(\dfrac{2016}{2^{2015}}+\dfrac{2017}{2^{2016}}\)
so sánh T với 3
Bạn Phong Thần trả lời hay quá.
uk, cái bạn tên Phong Thần công nhận giỏi thật nha
Cho tổng T = \(\dfrac{2}{2^1}\)+\(\dfrac{3}{2^2}\)+\(\dfrac{4}{2^3}\)+...+\(\dfrac{2016}{2^{2015}}\)+\(\dfrac{2017}{2^{2016}}\)
So sánh T với 3
Tính : A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}}\)
Ta có:
\(\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}\)
\(=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{1}{2016}\)
\(=1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{1}{2016}\right)\)
\(=\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2016}+\dfrac{2017}{2017}\)
\(=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
Do đó: \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\right)}=\dfrac{1}{2017}\)
Vậy...
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Rút gọn:
(\(\dfrac{2016}{1}+\dfrac{2015}{2}+...+\dfrac{2}{2015}+\dfrac{1}{2016}\)) : (\(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\))
đặt phân số trên là A
tử là
(1+2015/2)+...+(1+2/2015)+(1+1/2016)+1
=2017/2+....+2017/2015+2017/2016+2017/2017
=2017.(1/2+...+1/2015+1/2016+1/2017)
=>A=\(\dfrac{2017.\left(\dfrac{1}{2}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
Vậy A=2017
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Cho \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2017}{2^{2016}}+\dfrac{2018}{2^{2017}}\) . So sánh T và 3
So sánh \(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\) và \(B=2018\)
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
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\(A=\dfrac{2016^2+1^2}{2016\cdot1}+\dfrac{2015^2+2^2}{2015\cdot1}+\dfrac{2014^2+3^2}{2014\cdot3}+...+\dfrac{1009^2+1008^2}{1009\cdot1008}\)
và \(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\)Tìm A/B
s=\(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{n+1}{2^n}+...+\dfrac{2016}{2^{2015}}\)
xét tổng S gồm 2015 số hạng sau . Hay so sánh S với 3
tìm x biết (\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{2018}\)).x=\(\dfrac{2017}{1}+\dfrac{2016}{2}+\dfrac{2015}{3}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)
\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)
\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)
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