# Violympic toán 7

CM:

$\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{4}$

27 tháng 10 2017 lúc 19:29

Xét thừa số tổng quát:

$\dfrac{1}{t^3}< \dfrac{1}{t^3-t}=\dfrac{1}{t\left(t^2-1\right)}=\dfrac{1}{t\left(t+1\right)\left(t-1\right)}=\dfrac{1}{\left(t-1\right)t\left(t+1\right)}$

Thay vào bài toán:

$\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{\left(2-1\right)2\left(2+1\right)}+\dfrac{1}{\left(3-1\right)3\left(3+1\right)}+\dfrac{1}{\left(4-1\right)4\left(4+1\right)}+....+\dfrac{1}{\left(2017-1\right)2017\left(2017+1\right)}=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2016.2017.2018}$

Đặt:$A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2016.2017.2018}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2016.2017}-\dfrac{1}{2017.2018}\right)=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2017.2018}\right)=\dfrac{1}{1.2.2}-\dfrac{1}{2.2017.2018}=\dfrac{1}{4}-\dfrac{1}{2.2017.2018}< \dfrac{1}{4}\left(đpcm\right)$

Bình luận (0)
27 tháng 10 2017 lúc 19:41

Ta có: $\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2015.2016.2017}$

$=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2015.2016.2017}\right)$

$=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\right)$

$=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\right)$

$=\dfrac{1}{4}-\dfrac{1}{2.2016.2017}< \dfrac{1}{4}$

$\Rightarrowđpcm$

Bình luận (10)

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