chứng minh : A = 1 / 1×2\(+\)1/2×3 \(+\)1/3×4\(+\).....\(+\)1/n(n+1)=n/n+1(n\(\in\)N*)
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MD
22 tháng 9 2017
Ta có:\(\dfrac{1}{2^3}< \dfrac{1}{1.2.3};\dfrac{1}{3^3}< \dfrac{1}{2.3.4};\dfrac{1}{4^3}< \dfrac{1}{3.4.5};...;\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Vậy:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Ta có:\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n.\left(n+1\right)}\right)\)=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{n.\left(n+1\right)}\right)\)
=\(\dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
Vì:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) hay \(A< \dfrac{1}{4}\)
NT
26 tháng 9 2017
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
6 tháng 3 2021
\(\Rightarrow A=2^{2n}-1=4^n-1=\left(4-1\right)\left(4^{n-1}+4^{n-2}+...+4+1\right)=3\cdot\left(4^{n-1}+4^{n-2}+...+4+1\right)⋮3\forall n\in N\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
\(A=\frac{n}{n+1}\)
\(\Leftrightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
A = \(1-\frac{1}{n+1}+0+0+...+0\)
A = \(\frac{n+1-1}{n+1}\)
A = \(\frac{n}{n+1}\left(đpcm\right)\)