Question

tính tổng cấp số nhân vô hạn \(\dfrac{1}{2};-\dfrac{1}{4};\dfrac{1}{8};...;\dfrac{\left(-1\right)^{n+1}}{2^n};...\)

Answer 1

\(\left\{{}\begin{matrix}u_1=\dfrac{1}{2}\\q=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow S=\dfrac{1}{2}.\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{1}{3}\)