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
\(A=1+4+4^2+4^3+...+4^{99}\\ \Rightarrow4A=4+4^2+4^3+...+4^{100}\\ \Rightarrow3.A=4^{100}-1\\ \Rightarrow A=\dfrac{4^{100}-1}{3}< \dfrac{4^{100}}{3}=\dfrac{B}{3}\\ \Rightarrow A< \dfrac{B}{3}\)
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 \(B-A=\dfrac{1}{2}\)
A=1+3+32+33+...+320A=1+3+32+33+...+320
⇒3A=3+32+33+34+...+320+321⇒3A=3+32+33+34+...+320+321
⇒3A−A=(3+32+33+34+...+321)−(1+3+32+33+...+320)⇒3A−A=(3+32+33+34+...+321)−(1+3+32+33+...+320)⇒2A=321−1⇒2A=321−1
⇒A=321−12=3212−12⇒A=321−12=3212−12
Mà ta có:
B=3212B=3212
⇒B−A=3212−(3212−12)=3212−3212+12=12⇒B−A=3212−(3212−12)=3212−3212+12=12
Vậy B−A=12
