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

User Pllx
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1 Answer

1 vote

Answer:
\bold{-(8)/(3)}

Explanation:


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

User RKP
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