chứng tỏ A < 2019.Biết A=2018+1/16+1/25+1/36+.....+1/100+1/121
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12 tháng 5 2022
Đặt \(D=1+4+...+4^{2019}\)
\(\Leftrightarrow4D=4+4^2+...+4^{2020}\)
\(\Leftrightarrow D=\dfrac{4^{2020}-1}{3}\)
\(C=75\cdot D+25\)
\(=25\left(4^{2020}-1\right)+25=25\cdot4\cdot4^{2019}⋮100\)
TH
3
OY
27 tháng 5 2021
A=1/22+1/32+...+1/92
Ta có:1/22>1/2.3,1/32>1/3.4,...,1/92>1/9.10
⇒A>1/2.3+1/3.4+...+1/9.10
A>1/2-1/3+1/3-1/4+...+1/9-1/10
A>1/2-1/10
A>2/5(đpcm)
MH
19 tháng 4 2022
a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)
\(24n+384=25n+325\)
\(25n-24n=384-325\)
\(n=59\)
LH
3
KN
9 tháng 8 2020
\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)
NH
9 tháng 8 2020
\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\)( đpcm )
NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 3 2019
\(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A< \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{1}{2}-\frac{1}{100}=\frac{49}{100}=\left(\frac{7}{10}\right)^2\)
Ta có \(\frac{25}{36}=\left(\frac{5}{6}\right)^2\)
Ta thấy \(\frac{5}{6}=\frac{25}{30}>\frac{7}{10}=\frac{21}{30}\Rightarrow\left(\frac{7}{10}\right)^2< \left(\frac{5}{6}\right)^2\Rightarrow A< \left(\frac{7}{10}\right)^2< \left(\frac{5}{6}\right)^2=\frac{25}{36}\)
NT
1
10 tháng 7 2016
Theo đề bài thì A không phải là số tự nhiên suy ra A<1. Ta có
1/4+1/9+1/16+1/25+...+1/100
=1/2^2+1/3^2+1/4^2+1/5^2+...+1/10^2
=1/2x2+1/3x3+1/4x4+1/5x5+...+1/10x10<1/1x2+1/2x3+1/3x4+1/4x5+...+1/9x10
=>A<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
=>A<1/1-1/10
=>A<9/10
Vì 9/10<1=>A<1
Vậy A không phải số tự nhiên
ko b
Ta có:
\(\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{36}+...+\dfrac{1}{100}+\dfrac{1}{121}\\ =\dfrac{1}{4\times4}+\dfrac{1}{5\times5}+\dfrac{1}{6\times6}+...+\dfrac{1}{10\times10}+\dfrac{1}{11\times11}\\ < \dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+...+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)
Do \(\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{36}+...+\dfrac{1}{100}+\dfrac{1}{121}< \dfrac{1}{6}\)
\(\Rightarrow A< 2018+\dfrac{1}{6}< 2018+1=2019\)
Vậy \(A< 2019\)