Question

Cho A=1+3+3²+...+3²⁰ và B = 3²¹:2

 Tính B-A

Answer 1

\(3A=3+3^2+3^3+....+3^{21}\Leftrightarrow3A-A=2A=3^{21}-1\Rightarrow A=\frac{3^{21}-1}{2}\)

\(B-A=\frac{3^{21}}{2}-\frac{3^{21}-1}{2}=\frac{1}{2}\)

Answer 2

1/2 nha hi hi hi hi 

Answer 3

1/2

Answer 4

Nhân cả 2 vế của A với 3 ta được ;

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

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

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

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

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

Ta có : B-A=\(3^{21}\):2-\(\dfrac{3^{21}-1}{2}\)=\(\dfrac{3^{21}}{2}\)-\(\dfrac{3^{21}-1}{2}\)=\(\dfrac{3^{21}-3^{21}+1}{2}\)=\(\dfrac{1}{2}\)

Answer 5

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

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

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

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

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

Answer 6

3
\Rightarrow3A=3+3^2+3^3+...+3^{21}⇒3A=3+32+33+...+321
\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{21}\right)-\left(1+3+3^2+3^3+...+3^{20}\right)⇒3A−A=(3+32+33+...+321)−(1+3+32+33+...+320)
\Rightarrow2A=3^{21}-1⇒2A=321−1
\Rightarrow A=\frac{3^{21}-1}{2}=\frac{3^{21}}{2}-\frac{1}{2}⇒A=2321−1​=2321​−21​
Ta lại có:
B=\frac{3^{21}}{2}B=2321​
\Rightarrow B-A=\left(\frac{3^{21}}{2}-\frac{1}{2}\right)-\frac{3^{21}}{2}=\frac{1}{2}⇒B−A=(2321​−21​)−2321​=21​