Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)
\(2B=1-\frac{1}{729}\)
\(B=\frac{1-\frac{1}{729}}{2}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2C-C=\left(1+\frac{1}{2}+...+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)\)
\(C=1-\frac{1}{64}\)
A={\(\frac{1}{x^2}\)x={1;2;3;.....;15}}
B={x+a I a={1;3;5;7;.....} ; x2=a1+x1;x3=a3+x3....}