Question

[Phùng Nguyệt Minh]

1 + 1/2 + 1/4 + 1/8 +...+ 1/4096 + 1/8192

Answer 1

[Lung Thị Linh]

Đặt tổng trên = A

\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{4096}+\frac{1}{8192}\)

\(A.2=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}+\frac{1}{4096}\)

\(A.2-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}+\frac{1}{4096}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{4096}+\frac{1}{8192}\right)\)

\(A=2-\frac{1}{8192}=\frac{16383}{8192}\)
 

Answer 2

[soyeon_Tiểu bàng giải]

Đặt A = 1 + 1/2 + 1/4 + 1/8 + ... + 1/4096 + 1/8192

2A = 2 + 1 + 1/2 + 1/4 + ... + 1/2048 + 1/4096

2A - A = (2 + 1 + 1/2 + 1/4 + ... + 1/2048 + 1/4096) - (1 + 1/2 + 1/4 + 1/8 +... + 1/4096 + 1/8192)

A = 2 - 1/8192

A = 16383/8192