∑k=0n4−k\sum_{k=0}^{n}{4^{-k}}k=0∑n4−k
asked by guest on Nov 28, 2024 at 5:59 am
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MathBot Answer:
∑k=0n4−k=43−13⋅4n\sum_{k=0}^{n} 4^{- k} = \frac{4}{3} - \frac{1}{3 \cdot 4^{n}}k=0∑n4−k=34−3⋅4n1
∑k=0n4−k=(−(14)1+n+1)−14+1 by the identity∑n=0man=1−am+11−a=43−13⋅4n\begin{aligned}\sum_{k=0}^{n} 4^{- k} &= \frac{\left(- \left(\frac{1}{4}\right)^{1 + n} + 1\right)}{\frac{-1}{4} + 1} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} a^n = \frac{1-a^{m+1}}{1-a}}\\&= \frac{4}{3} - \frac{1}{3 \cdot 4^{n}}\end{aligned}k=0∑n4−k=4−1+1(−(41)1+n+1) by the identityn=0∑man=1−a1−am+1=34−3⋅4n1
