a: Sửa đề: \(1\frac12+2\frac14+3\frac18+4\frac{1}{16}+5\frac{1}{32}+6\frac{1}{64}\)
=(1+2+3+4+5+6)+\(\left(\frac12+\frac14+\cdots+\frac{1}{64}\right)\)
=21+\(\left(\frac12+\frac14+\cdots+\frac{1}{64}\right)\)
Đặt \(A=1\frac12+2\frac14+3\frac18+4\frac{1}{16}+5\frac{1}{32}+6\frac{1}{64}\)
Đặt \(B=\frac12+\frac14+\cdots+\frac{1}{64}\)
=>2 B=\(1+\frac12+\cdots+\frac{1}{32}\)
=>2B-B=\(1+\frac12+\cdots+\frac{1}{32}-\frac12-\frac14-\cdots-\frac{1}{64}\)
=>B=\(1-\frac{1}{64}=\frac{63}{64}\)
\(A=21+\left(\frac12+\frac14+\cdots+\frac{1}{64}\right)\)
\(=21+\frac{63}{64}=\frac{1407}{64}\)
b: Đặt \(C=\frac12+\frac14+\frac18+\cdots+\frac{1}{1024}\)
=>\(2\times C=1+\frac12+\frac14+\cdots+\frac{1}{512}\)
=>\(2\times C-C=1+\frac12+\frac14+\cdots+\frac{1}{512}-\frac12-\frac14-\cdots-\frac{1}{1024}\)
=>\(C=1-\frac{1}{1024}=\frac{1023}{1024}\)
