Question

\(\frac{1}{2}\)+  \(\frac{1}{4}\)+  \(\frac{1}{8}\)+  \(\frac{1}{16}\)+ \(\frac{1}{32}\)+  \(\frac{1}{64}\)=  ??????????????

Đề bài : Hãy tính tổng các phân số dưới đây.

 

Answer 1

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

\(A=1-\frac{1}{64}=\frac{63}{64}\)

Answer 2

Ta thấy: 1/2 = 1 - 1/2

1/2 + 1/4 = 3/4 = 1- 1/4

1/2 + 1/4 + 1/8 = 7/8 = 1 - 1/8

Tương tự ta có:

1/2 + 1/4 +1/8 + 1/16 + 1/32 + 1/64 = 1 - 1/64 = 63/64

Answer 3

63/64 nha bạn, học tốt nha

Answer 4

28/64+16/64+6/64+4/64+2/64+1/64

=28+16+6+4+2+1/64

=57/64

Answer 5

63/64