Question

find the sum of the finite geometric series.
\\(\sum_{n=1}^{5} 4^n\\)

Explanation:

Step1: Recall the formula for the sum of a finite geometric series

The sum of a finite geometric series \(\sum_{n = 1}^{k} ar^{n - 1}\) is given by \(S_{k}=\frac{a(1 - r^{k})}{1 - r}\) when \(r
eq1\). However, our series is \(\sum_{n = 1}^{5}4^{n}\), which can be rewritten as \(\sum_{n = 1}^{5}4\times4^{n - 1}\). So here, \(a = 4\) (the first term), \(r = 4\) (the common ratio), and \(k = 5\) (the number of terms).

Step2: Apply the sum formula

Using the formula \(S_{k}=\frac{a(r^{k}-1)}{r - 1}\) (since \(r>1\), this form is more convenient to avoid negative denominators), we substitute \(a = 4\), \(r = 4\), and \(k = 5\) into the formula.

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Or we can also calculate the terms directly: when \(n = 1\), \(4^{1}=4\); \(n = 2\), \(4^{2}=16\); \(n = 3\), \(4^{3}=64\); \(n = 4\), \(4^{4}=256\); \(n = 5\), \(4^{5}=1024\). Then sum them up: \(4 + 16+64 + 256+1024=1364\).

Answer:

\(1364\)