infinite series - 2

[logistic_guy]

Solve.

\(\displaystyle \sum_{k=0}^{\infty}3\left(\frac{1}{5}\right)^k\)

 

[khansaheb]

Solve.

\(\displaystyle \sum_{k=0}^{\infty}3\left(\frac{1}{5}\right)^k\)

factor out '3' and you have a geometric series....

 

[logistic_guy]

\(\displaystyle \sum_{k=0}^{\infty}3\left(\frac{1}{5}\right)^k = \frac{3}{1 - \frac{1}{5}} = \frac{15}{4} = \textcolor{indigo}{3.75}\)