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ND
27 tháng 2 2018
\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\) \(\Rightarrow2S=1-\dfrac{1}{2n+1}\)
\(\Rightarrow S=\dfrac{n}{2n+1}\)
27 tháng 2 2018
Ta có : \(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
ta được \(\dfrac{1}{1.3}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right);\dfrac{1}{3.5}=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right);\dfrac{1}{5.7}=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\) vậy \(S=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}\)
31 tháng 3 2018
ta có (a-1)2 ≥ 0 ∀a
<=> a2-2a+1 ≥ 0
<=>a2+4a-2a+1 ≥ 4a (cộng cả 2 vế va 4a)
<=> a2+2a+1 ≥ 4a
<=> (a+1)2 ≥ 4a
CM tương tự ta đc
(b+1)2 ≥ 4b
(c+1)2 ≥ 4c
Nhân các vế với nhau ta có
[(a+1)2+(b+1)2 +(c+1)2 ]2 ≥ 4a.4b.4c
<=> [(a+1)2+(b+1)2 +(c+1)2 ]2 ≥64abc
<=> [(a+1)2+(b+1)2 +(c+1)2 ]2 ≥64 (vì abc =1)
<=> (a+1)2+(b+1)2 +(c+1)2 ≥8 (đpcm)
21 tháng 6 2016
bạn làm theo công thức \(\frac{n}{n.\left(n+1\right)}=\frac{n}{n}-\frac{n}{n+1}\)
TM
21 tháng 6 2016
a)Đặt A= \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(\Rightarrow2A=1-\frac{1}{2n+1}< 1\)
\(\Rightarrow A< \frac{1}{2}\)(đpcm)
b)Ta có: \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1+1-\frac{1}{n}\)
\(=2-\frac{1}{n}< 2\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}< 2\)
\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 2\)(đpcm)
5 tháng 6 2022
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
7 tháng 6 2022
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)
\(=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
NT
0
PN
1
15 tháng 4 2020
22n(22n+1-1)-1
\(=2^{4n+1}-2^{2n}-1=2.2^{4n}-2^{2n}-1\)
\(=2\left(2^{2n}\right)^2-2^{2n}-1=A\)
Đặt \(2^{2n}=t\)
\(\Rightarrow A=2t^2-t-1=\left(2t+1\right)\left(t-1\right)\)
\(=\left(2.2^{2n}+1\right)\left(2^{2n}-1\right)\)
\(=\left(2^{2n+1}+1\right)\left(2^{2n}-1\right)=\left(2+1\right)\left(2^{2n}-2^{2n-1}+...+1\right)\left(2+1\right)\left(2^{2n-1}+...-1\right)\)
\(=9.B\)
Vậy \(A⋮9\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
=\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
=\(\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
=\(\dfrac{1}{2}-\dfrac{1}{4n+2}< \dfrac{1}{2}\)
đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=> 2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
<=> 2A=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{2n-2}-\dfrac{1}{2n+1}\)
<=>2A=\(1-\dfrac{1}{2n+1}\)
<=> A=\(\left(1-\dfrac{1}{2n+1}\right)\)\(.\dfrac{1}{2}\)
<=> A=\(\dfrac{1}{2}-\dfrac{1}{2\left(2n+1\right)}\)
=>\(A< \dfrac{1}{2}\) (đpcm)