sum(2^k, k=1, n) — step-by-step solution

Let's evaluate:

\sum_{k = 1}^{n} 2^{k}

For

\sum_{K = a}^{b} c^{(K + d)}

, the following formula applies:

c^{(a + d)} \cdot \frac{c^{(b - a + 1)} - 1}{c - 1}

2 \cdot \frac{2^{(n - 1 + 1)} - 1}{2 - 1}

ℹ

2 \cdot \frac{2^{(n - 1 + 1)} - 1}{2 - 1}

\frac{2 \cdot (2^{(n - 1 + 1)} - 1)}{2 - 1}

‖

- 1 + 1

0

‖

2 - 1

1

\frac{2 \cdot (2^{(n + 0)} - 1)}{1}

▷Result: $2 \cdot (2^{n} - 1)$