ai làm cho 2 k

Đặt \(A=\frac15+\frac{2}{5^2}+\cdots+\frac{2016}{5^{2016}}\)

=>\(5A=1+\frac25+\cdots+\frac{2016}{5^{2015}}\)

=>\(5A-A=1+\frac25+\cdots+\frac{2016}{5^{2015}}-\frac15-\frac{2}{5^2}-\cdots-\frac{2016}{5^{2016}}\)

=>\(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

Đặt \(B=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}\)

=>\(5B=1+\frac15+\cdots+\frac{1}{5^{2014}}\)

=>\(5B-B=1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}\)

=>\(4B=1-\frac{1}{5^{2015}}=\frac{5^{2015}-1}{5^{2015}}\)

=>\(B=\frac{5^{2015}-1}{4\cdot5^{2015}}\)

TA có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

\(=1+\frac{5^{2015}-1}{4\cdot5^{2015}}-\frac{2016}{5^{2016}}=1+\frac{5^{2016}-5-8064}{4\cdot5^{2016}}=1+\frac14-\frac{8069}{4\cdot5^{2016}}\)

=>\(4A<1+\frac14=\frac54\)

=>\(A<\frac{5}{16}\)

mà \(\frac{5}{16}<\frac{5}{15}=\frac13\)

nên \(A<\frac13\) (1)

Ta có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

=>\(20A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}\)

=>\(20A-4A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}-1-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

=>\(16A=5-\frac{2017}{5^{2015}}-\frac{2016}{5^{2016}}>5\)

=>\(A>\frac{5}{16}\)

=>\(A>\frac{4}{16}=\frac14\) (2)

Từ (1),(2) suy ra 1/4<A<1/3

giúp mình với

= 2 mũ 2018 - 2 mũ 2016

=2 mũ 2016 . 2 mũ 2 - 2 mũ 2016

=2 mũ 2016.(2 mũ 2 - 1)

=2 mũ 2016.(4 -1 )

=(2 mũ 2016.3) chia hết cho 3

Vậy:2 mũ 2018 - 2 mũ 2016 chia hết cho 3

c:lẻ=> x+2017:chẵn chia hết cho 2 
vậy a chia hết cho 2 
Nếu x :chẵn => x+2016:chẵn chia hết cho 2 
vậy a :2 
Kết luận : x thuộc N thì a chia hết cho 2 
kết mk nha ^^

Ta có : 2²⁰¹⁶ = 2²⁰¹³.2³ = 2²⁰¹³.8

Lại có : 2²⁰¹⁵ + 2²⁰¹⁴ + 2²⁰¹³ = 2²⁰¹³.(2² + 2 + 1) = 2²⁰¹³.7 

Vì 7 < 8 

=>  2²⁰¹³.8 <  2²⁰¹³.7

=> 2²⁰¹⁶ < 2²⁰¹⁵ + 2²⁰¹⁴ + 2²⁰¹³ = 2²⁰¹³ (đpcm)