Question

Tính \(lim\dfrac{1+2+4+...+2^n}{1+5+25+...+5^n}\)

Answer 1

Đặt A=1+2+4+...+2^n

=>2A=2+2^2+2^3+...+2ⁿ⁺¹

=>\(A=2^{n+1}-1\)

Đặt B=1+5+5^2+...+5^n

=>\(5B=5+5^2+5^3+...+5^{n+1}\)

=>\(4B=5^{n+1}-1\)

=>\(B=\dfrac{5^{n+1}-1}{4}\)

\(lim\left(\dfrac{A}{B}\right)=\lim\limits\dfrac{2^{n+1}-1}{\dfrac{5^{n+1}-1}{4}}=\lim\limits\dfrac{2^{n+3}-4}{5^{n+1}-1}\)

\(=\lim\limits\dfrac{2^n\cdot8-4}{5^n\cdot5-1}\)

\(=\lim\limits\dfrac{\left(\dfrac{2}{5}\right)^n\cdot8-\dfrac{4}{5^n}}{5-\dfrac{1}{5^n}}=\dfrac{0}{5}=0\)