Question

[Trần Minh Quân]

giải ra hộ mik nhé

Answer 1

[Nguyễn Lê Phước Thịnh]

Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)

=>\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}+\dfrac{1}{3^7}\)

=>\(3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^5}+\dfrac{1}{3^6}\)

=>\(3A-A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^5}+\dfrac{1}{3^6}-1-\dfrac{1}{3}-...-\dfrac{1}{3^7}\)

=>\(2A=3-\dfrac{1}{3^7}=\dfrac{3^8-1}{3^7}\)

=>\(A=\dfrac{3^8-1}{3^7\cdot2}\)