Lê Quốc Trần Anh Coordinator
13/07/2017 at 12:08-
LIVERPOOL 13/07/2017 at 16:42A= \(-\dfrac{1}{3}+\dfrac{1}{3^2}-...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)
=>3A=\(-1+\dfrac{1}{3}-...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
=>4A=\(-1-\dfrac{1}{3^{101}}\) =>A=\(\dfrac{-1-\dfrac{1}{3^{101}}}{4}\)
Lê Quốc Trần Anh selected this answer. -
FA KAKALOTS 28/01/2018 at 22:13A= −13+132−...+13100−13101
=>3A=−1+13−...+1399−13100
=>4A=−1−13101
=>A=−1−131014