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

Let's calculate:

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

For

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

, you can use the formula:

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

ⅰ

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

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

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

ⓘ Multiplying a fraction by a number multiplies its numerator:

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

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

■Calculate the exponent before evaluating the power:

- 1 + 1

0

■The numerator and the denominator are computed before the division:

2 - 1

1

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

ⓘ Adding 0 changes nothing:

n + 0

n

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

ⓘ Multiply each term of the sum by the other factor, using the distributive law:

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

2^{n + 1} - 2

▶Result:

2^{n + 1} - 2