Question

1. Tính tổng:

B = 2 - 4 - 6 + 8 + 10 - 12 - 14 + 16 + ... + 2002 - 2004 - 2006 + 2008

2. Cho A = 1 + 3 + 3² + 3³ + ... + 3²⁰, B = 3²¹ : 2

Tính B - A

3. Cho A = 1 + 4 + 4² + 4³ + ... + 4⁹⁹, B = 4¹⁰⁰

Chứng minh rằng A < \(\dfrac{B}{3}\)

Answer 1

1. Tính tổng:

B = 2 - 4 - 6 + 8 + 10 - 12 - 14 + 16 + ... + 2002 - 2004 - 2006 + 2008

=> ( 2 - 4 - 6 + 8 )+ (10 - 12 - 14 + 16) + ... + (2002 - 2004 - 2006 + 2008)

=> (-8+ 8) +(-16+ 16) +.........+ ( -2008+ 2008)(1)

=> 0+0+...........+0

=> 0

Ta thấy (1) đều là những số đối nên kết quả đường nhiên bằng 0

Answer 2

\(A=1+4+4^2+4^3+...+4^{99}\\ \Rightarrow4A=4+4^2+4^3+...+4^{100}\\ \Rightarrow3.A=4^{100}-1\\ \Rightarrow A=\dfrac{4^{100}-1}{3}< \dfrac{4^{100}}{3}=\dfrac{B}{3}\\ \Rightarrow A< \dfrac{B}{3}\)

Answer 3

2.
\(A=1+3+3^2+3^3+...+3^{20}\)
\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{20}+3^{21}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+3^4+...+3^{21}\right)-\left(1+3+3^2+3^3+...+3^{20}\right)\)\(\Rightarrow2A=3^{21}-1\)
\(\Rightarrow A=\dfrac{3^{21}-1}{2}=\dfrac{3^{21}}{2}-\dfrac{1}{2}\)
Mà ta có:
\(B=\dfrac{3^{21}}{2}\)
\(\Rightarrow B-A=\dfrac{3^{21}}{2}-\left(\dfrac{3^{21}}{2}-\dfrac{1}{2}\right)=\dfrac{3^{21}}{2}-\dfrac{3^{21}}{2}+\dfrac{1}{2}=\dfrac{1}{2}\)
Vậy \(B-A=\dfrac{1}{2}\)

Answer 4

Mà ta có:


Vậy