Question

\sum _{n=1}^{\infty }−4\left(\frac{−1}{2}\right)^{n-1}

Answer 1

Answer: $\bold{-\dfrac{8}{3}}$

Step-by-step explanation:

$\sum \limits_{n=1}^{\infty}-4\bigg(\dfrac{-1}{2}\bigg)^{n-1}\implies a_1=-4, r=-\dfrac{1}{2}\\\\\\\text{Use the formula for the sum of an infinite geometric series:}\\S=\dfrac{a_1}{1-r}\\\\\\.\ =\dfrac{-4}{1-(-\frac{1}{2})}\\\\\\.\ =\dfrac{-4}{\frac{3}{2}}\\\\\\.\ =-4\times \dfrac{2}{3}\\\\\\.\ =\large\boxed{-\dfrac{8}{3}}$