Question

A = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^{100}}\)

B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}\)

mk cần gấp

Answer 1

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(A=2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)\(\Rightarrow A=1-\dfrac{1}{2^{100}}\)

Answer 2

MoMo Nguyễn Thị Hồng Nhung Trần Thọ Đạt giúp mk vs

Answer 3

Đề là chứng minh hay tínhHoàng Quỳnh Trang

Answer 4

đề bài là tính tổng nha

C =

\(\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+\dfrac{3}{1+2+3+4}+...+\dfrac{3}{1+2+3+....+100}\)