Evaluate sum from n=3 to 12 of 20(0.5)^(n-1)

$\sum_{n = 3}^{12} 20 {(0.5)}^{n - 1}$

Split the summation to make the starting value of $n$ equal to $1$ .

$\sum_{n = 3}^{12} 20 {(0.5)}^{n - 1} = \sum_{n = 1}^{12} 20 {(0.5)}^{n - 1} - \sum_{n = 1}^{2} 20 {(0.5)}^{n - 1}$

Evaluate $\sum_{n = 1}^{12} 20 {(0.5)}^{n - 1}$ .

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The formula for the geometric series is:

$\sum_{k = 1}^{n} a r^{n - 1} = a + a r + a r^{2} + \dots + a r^{n - 1} = a (\frac{1 - r^{n}}{1 - r})$

Substitute the values into the formula and make sure to multiply by the front term.

$20 (\frac{1 - {(0.5)}^{12}}{1 - 0.5})$

Simplify.

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Simplify the numerator.

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Raise $0.5$ to the power of $12$ .

$20 \frac{1 - 1 \cdot 0.00024414}{1 - 0.5}$

Multiply $- 1$ by $0.00024414$ .

$20 \frac{1 - 0.00024414}{1 - 0.5}$

Subtract $0.00024414$ from $1$ .

$20 \frac{0.99975585}{1 - 0.5}$

Reduce the expression by cancelling the common factors.

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Subtract $0.5$ from $1$ .

$20 (\frac{0.99975585}{0.5})$

Cancel the common factor of $0.5$ .

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Factor $0.5$ out of $20$ .

$0.5 (40) \frac{0.99975585}{0.5}$

Cancel the common factor.

$0.5 \cdot 40 \frac{0.99975585}{0.5}$

Rewrite the expression.

$40 \cdot 0.99975585$

Multiply $40$ by $0.99975585$ .

$39.99023437$

Evaluate $\sum_{n = 1}^{2} 20 {(0.5)}^{n - 1}$ .

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Expand the series for each value of $n$ .

$20 {(0.5)}^{(1) - 1} + 20 {(0.5)}^{(2) - 1}$

Simplify.

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Simplify each term.

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Subtract $1$ from $1$ .

$20 \cdot {0.5}^{0} + 20 {(0.5)}^{(2) - 1}$

Anything raised to $0$ is $1$ .

$20 \cdot 1 + 20 {(0.5)}^{(2) - 1}$

Multiply $20$ by $1$ .

$20 + 20 {(0.5)}^{(2) - 1}$

Subtract $1$ from $2$ .

$20 + 20 \cdot {0.5}^{1}$

Evaluate the exponent.

$20 + 20 \cdot 0.5$

Multiply $20$ by $0.5$ .

$20 + 10$

Add $20$ and $10$ .

$30$

Replace the summations with the values found.

$39.99023437 - 30$

Subtract $30$ from $39.99023437$ .

$9.99023437$

Evaluate sum from n=3 to 12 of 20(0.5)^(n-1)

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