Question

[Hoàng Thị Kim Chi]

Cho A= 1+3+3²+3³+...+3¹⁰⁰

B= 3¹⁰⁰ : 2.Tính B-A

Answer 1

[OoO_TNT_OoO]

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

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

\(3A=3+3^2+3^3+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-1\)

Answer 2

[không có tên]

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

\(3A=3+3^2+...+3^{101}\)

\(3a-a=(3+3^2+...+3^{101}-(1+3+3^2+...+2^{100})\)

\(2A=3^{101}-1\)

\({A=2^{101}-1}/{2}\)

\(=> B-A = 3^{100}/2 - 3^{101}-1/2\)