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A=1*2+2*3+3*4+...+2017*2018
3A=1*2*3+2*3*(4-1)+...+2017*2018*(2019-2016)
3A=1*2*3+2*3*4-1*2*3+...+2017*2018*2019-2016*2017*2018
3A=2017*2018*2019
A=\(\frac{2017.2018.2019}{3}\)
mk chỉ biết tính a thôi
Ta có:
A = 1 + 5 + 52 + 53 + 54 + ...+ 52017
A = \(\frac{5^{2017}-1}{5-1}\)
B = \(\frac{5^{2018}-1}{2-1}\)
=> \(4A=\frac{5^{2017}-1}{4}.4=5^{2017}-1< B=5^{2018}-1\)
Vậy 4A < B
Ta có: 5A=5(1+5+52+....+52017)
5A=5+52+53+....+52018
5A-A=(5+52+53+...+52018)-(1+5+52+....+52017)
4A=52018-1
Vì 4A=52018-1. Mà 52018-1=52018-1
Suy ra:4A=B
\(3A=3.\left(1+3+3^2+3^3+...+3^{2017}\right)\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=3+3^2+3^3+...+3^{2018}-\left(1+3+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}< \frac{3^{2018}}{2}=B=>A< B\)
\(A=1+3+3^2+3^3+....+3^{2017}.\)
\(\Rightarrow3A=3+3^2+3^3+3^4+......+3^{2018}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+3^4+....+3^{2018}\right)-\left(1+3+3^2+3^3+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(\Rightarrow A=\frac{3^{2018}-1}{2}< \frac{3^{2018}}{2}\)
\(\Rightarrow A< B\)
1 Ta có: 201810 + 20189 = 20189.(2018 + 1) = 20189. 2019
201710 = 20179.2017
=> 201810 + 20189 > 201710
2. A = 1 + 2 + 22 + 23 + ... + 2100
2A = 2(1 + 2 + 22 + 23 + ... + 2100)
2A = 2 + 22 + 23 + ... + 2101
2A - A = (2 + 22 + 23 + ... + 2101) - (1 + 2 + 22 +. ... + 2100)
A = 2101 - 1
B = 1 + 6 + 11 + 16 + ... + 51
B = (51 + 1)[(51 - 1) : 5 + 1] : 2
B = 52. 11 : 2
B = 286
\(\frac{2015}{2018^3}-\frac{2017}{2018^3}=-\frac{2}{2018^3}\) \(\frac{2015}{2018^4}-\frac{2017}{2018^4}=-\frac{2}{2018^4}\)
vì \(-\frac{2}{2018^3}< -\frac{2}{2018^4}\Rightarrow\frac{2015}{2018^3}-\frac{2017}{\cdot2018^3}< \frac{2015}{2018^4}-\frac{2017}{2018^4}\)
chuyển vế ta đc : \(\frac{2015}{2018^3}+\frac{2017}{2018^4}< \frac{2017}{2018^3}+\frac{2015}{2018^4}\)
A = 2015.2018/2018^4 + 2017/2018^4 = 2015.2018+2017/2018^4
B=2017.2018/2018^4 + 2015/2018^4 = 2017.2018+2015/2018^4
Vì 2015.2018+2017<2017.2018+2015 nên A<B
a>b
K CHO MÌNH NHÉ
cái này là khi chiều mới thi nầy
Giải:
Ta có:A=1.2+2.3+3.4+...+2017.2018
3A=1.2.3 2.3.3+...+2017.2018.3
=1.2.(3-0)+2.3.(4-1)+...+2017.2018.(2019-2016)
=1.2.3+2.3.4+...+2017.2018.2019-1.2.0-2.3.1-...-2017.2018.1016
=2017.2018.2019-1.2.0
=2017.2018.2019
=>A=2017.2018.2019/3=2018.(2017.2019)/3
Và B=20183/3=2018.2018.2018/3=2018.(2018.2018)/3
Lại có: 2017.2019=2017.(2018+1)=2017.2018+2017
2018.2018=(2017+1).2018=2017.2018+2018
Mà 2017.2018+2017<2017.2018+2018 =>2017.2019<2018.2018
=>2018.(2017.2019)<2018.(2018.2018)
=>A=2018.(2017.2019)/3<2018.(2018.2018)/3=B
=>A<B
Vậy A<B
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