Question

Cho A = 1 + 3 + 3^2 + .........+ 3^20 , B = 3^31 : 2

Tính B - A

Answer 1

Nếu đề đúng thì chắc không tính được

Sửa đề:

Cho \(A=1+3+3^2+...+3^{20}\). \(B=3^{21}\div2\)

Tính \(B-A\)

Giải:

Ta có:

\(A=1+3+3^2+...+3^{20}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{21}\)

\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{21}\right)-\left(1+3+3^2+...+3^{20}\right)\)

\(\Rightarrow2A=3^{21}-1\)

\(\Rightarrow A=\dfrac{3^{21}-1}{2}=\dfrac{3^{21}}{2}-\dfrac{1}{2}\)

Lại có:

\(B=3^{21}\div2=\dfrac{3^{21}}{2}\)

\(\Rightarrow B-A=\dfrac{3^{21}}{2}-\left(\dfrac{3^{21}}{2}-\dfrac{1}{2}\right)\)

\(\Rightarrow B-A=0-\dfrac{1}{2}\)

\(\Rightarrow B-A=\dfrac{-1}{2}\)

Vậy \(B-A=\dfrac{-1}{2}\)