Bài 1:
\(A=2^3+2^5+...+2^{19}\)
\(2^2A=2^2\left(2^3+2^5+...+2^{19}\right)\)
\(4A=2^5+2^7+...+2^{21}\)
\(4A-A=\left(2^5+2^7+...+2^{21}\right)-\left(2^3+2^5+...+2^{19}\right)\)
\(3A=2^{21}-2^3\Rightarrow A=\dfrac{2^{21}-8}{3}\)
Bài 2:
\(B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{19}}\)
\(3B=3\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{19}}\right)\)
\(3B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{18}}\)
\(3B-B=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{18}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{19}}\right)\)
\(2B=\dfrac{1}{3}-\dfrac{1}{3^{19}}\Rightarrow B=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{19}}}{2}\)
B = \(\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{19}}\)
=> 3B = \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{18}}\)
=> 3B - B = \(\dfrac{1}{3}-\dfrac{1}{3^{19}}\)
=> 2B = \(\dfrac{1}{3}-\dfrac{1}{3^{19}}\)
=> B = \(\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{19}}}{2}\)
@Hà Cao Thu
