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

ai trả lời giúp mình

1^3+2^3+4^3+5^3=(1+2+3+4+5)^2

TA CÓ:

A=3⁰+3+3²+3³+........+3¹¹

(3⁰+3+3²+3³)+....+(3⁸+3⁹+3¹⁰+3¹¹)

3(0+1+3+3²)+......+3⁸(0+1+3+3²) 

3.13+....+3⁸.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)

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Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

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