nguyendinhtuong wrote:
What is the value of \(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)?
A. Between \(\frac{1}{8}\) and \(\frac{1}{4}\)
B. Between \(\frac{1}{16}\) and \(\frac{1}{4}\)
C. Between \(\frac{1}{32}\) and \(\frac{1}{16}\)
D. Between \(\frac{1}{16}\) and \(\frac{1}{8}\)
E. Between \(\frac{1}{4}\) and \(\frac{1}{2}\)
Solution.\(\begin{split}
A&=&\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{10}{5^{11}}+\frac{11}{5^{12}}\\
5A&=\frac{1}{5}+&\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\\
5A-A&=\frac{1}{5}+&\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\\
4A&=B-&\frac{11}{5^{12}}
\end{split}\)
\(\begin{split}
B&=&\frac{1}{5}+&\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{10}}+\frac{1}{5^{11}}\\
5B&=1+&\frac{1}{5}+&\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{10}}\\
4B&=1-&\frac{1}{5^{11}}\\
\end{split}\)
\(\begin{split}
4B&=\frac{5^{11}-1}{5^{11}}\\
B&=\frac{5^{11}-1}{4 \times 5^{11}}
\end{split}\)
Hence we have
\(\begin{split}
4A&=B-\frac{11}{5^{12}}=\frac{5^{11}-1}{4 \times 5^{11}}-\frac{11}{5^{12}}\\
4A&=\frac{5(5^{11}-1)-44}{4 \times 5^{12}}\\
4A&=\frac{5^{12}-5-44}{4 \times 5^{12}}\\
4A&=\frac{5^{12}-49}{4 \times 5^{12}}\\
A&=\frac{5^{12}-49}{16 \times 5^{12}}\\
A&=\frac{1}{16} \times \frac{5^{12}-49}{5^{12}}\\
A&=\frac{1}{16}\bigg( 1- \frac{49}{5^{12}} \bigg )\\
\end{split}\)
Since \(1- \frac{49}{5^{12}} < 1 \implies A < \frac{1}{16}\).
It's clear that the answer is C.
Also \(2 \times 49 < 5^{12} \implies \frac{49}{5^{12}} < \frac{1}{2} \implies 1- \frac{49}{5^{12}} > 1- \frac{1}{2}=\frac{1}{2}\)
Hence \(A>\frac{1}{16} \times \frac{1}{2}=\frac{1}{32}\).