$$\sum_{k=0}^{n}{4^{-k}}$$
You asked:
Evaluate the expression: \(\sum_{k = 0}^{n} {{4}^{-k}}\)
MathBot Answer:
\[\sum_{k=0}^{n} 4^{- k} = \frac{4}{3} - \frac{4^{- n}}{3}\]
$$\sum_{k=0}^{n}{4^{-k}}$$
You asked:
MathBot Answer:
\[\sum_{k=0}^{n} 4^{- k} = \frac{4}{3} - \frac{4^{- n}}{3}\]