Evaluate the Summation sum from k=1 to 5 of 3(-5)^(k-1)

$\sum_{k = 1}^{5} 3 {(- 5)}^{k - 1}$

The sum of a finite geometric series can be found using the formula $a (\frac{1 - r^{n}}{1 - r})$ where $a$ is the first term and $r$ is the ratio between successive terms.

Find the ratio of successive terms by plugging into the formula $r = \frac{a_{k + 1}}{a_{k}}$ and simplifying.

Substitute $a_{k}$ and $a_{k + 1}$ into the formula for $r$ .

$r = \frac{3 {(- 5)}^{k + 1 - 1}}{3 {(- 5)}^{k - 1}}$

Simplify.

Cancel the common factor of $3$ .

Cancel the common factor.

$r = \frac{3 {(- 5)}^{k + 1 - 1}}{3 {(- 5)}^{k - 1}}$

Rewrite the expression.

$r = \frac{{(- 5)}^{k + 1 - 1}}{{(- 5)}^{k - 1}}$

Cancel the common factor of ${(- 5)}^{k + 1 - 1}$ and ${(- 5)}^{k - 1}$ .

Factor ${(- 5)}^{k - 1}$ out of ${(- 5)}^{k + 1 - 1}$ .

$r = \frac{{(- 5)}^{k - 1} {(- 5)}^{k + 0 - (k - 1)}}{{(- 5)}^{k - 1}}$

Cancel the common factors.

Multiply by $1$ .

$r = \frac{{(- 5)}^{k - 1} {(- 5)}^{k + 0 - (k - 1)}}{{(- 5)}^{k - 1} \cdot 1}$

Cancel the common factor.

$r = \frac{{(- 5)}^{k - 1} {(- 5)}^{k + 0 - (k - 1)}}{{(- 5)}^{k - 1} \cdot 1}$

Rewrite the expression.

$r = \frac{{(- 5)}^{k + 0 - (k - 1)}}{1}$

Divide ${(- 5)}^{k + 0 - (k - 1)}$ by $1$ .

$r = {(- 5)}^{k + 0 - (k - 1)}$

Add $k$ and $0$ .

$r = {(- 5)}^{k - (k - 1)}$

Simplify each term.

Apply the distributive property.

$r = {(- 5)}^{k - k - - 1}$

Multiply $- 1$ by $- 1$ .

$r = {(- 5)}^{k - k + 1}$

Subtract $k$ from $k$ .

$r = {(- 5)}^{0 + 1}$

Add $0$ and $1$ .

$r = {(- 5)}^{1}$

Evaluate the exponent.

$r = - 5$

Find the first term in the series by substituting in the lower bound and simplifying.

Substitute $1$ for $k$ into $3 {(- 5)}^{k - 1}$ .

$a = 3 {(- 5)}^{1 - 1}$

Simplify.

Subtract $1$ from $1$ .

$a = 3 {(- 5)}^{0}$

Anything raised to $0$ is $1$ .

$a = 3 \cdot 1$

Multiply $3$ by $1$ .

$a = 3$

Substitute the values of the ratio, first term, and number of terms into the sum formula.

$3 \frac{1 - {(- 5)}^{5}}{1 - - 5}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $5$ .

$3 \frac{1 - - 3125}{1 - - 5}$

Multiply $- 1$ by $- 3125$ .

$3 \frac{1 + 3125}{1 - - 5}$

Add $1$ and $3125$ .

$3 \frac{3126}{1 - - 5}$

Simplify the denominator.

Multiply $- 1$ by $- 5$ .

$3 \frac{3126}{1 + 5}$

Add $1$ and $5$ .

$3 (\frac{3126}{6})$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$3 \frac{3126}{3 (2)}$

Cancel the common factor.

$3 \frac{3126}{3 \cdot 2}$

Rewrite the expression.

$\frac{3126}{2}$

Divide $3126$ by $2$ .

$1563$

Evaluate the Summation sum from k=1 to 5 of 3(-5)^(k-1)

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