Evaluate the Summation sum from j=1 to 4 of 6(-2)^(j-1)

$\sum_{j = 1}^{4} 6 {(- 2)}^{j - 1}$

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 $r = \frac{a_{j + 1}}{a_{j}}$ and simplifying.

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

$r = \frac{6 {(- 2)}^{j + 1 - 1}}{6 {(- 2)}^{j - 1}}$

Simplify.

Cancel the common factor of $6$ .

Cancel the common factor.

$r = \frac{6 {(- 2)}^{j + 1 - 1}}{6 {(- 2)}^{j - 1}}$

Rewrite the expression.

$r = \frac{{(- 2)}^{j + 1 - 1}}{{(- 2)}^{j - 1}}$

Cancel the common factor of ${(- 2)}^{j + 1 - 1}$ and ${(- 2)}^{j - 1}$ .

Factor ${(- 2)}^{j - 1}$ out of ${(- 2)}^{j + 1 - 1}$ .

$r = \frac{{(- 2)}^{j - 1} {(- 2)}^{j + 0 - (j - 1)}}{{(- 2)}^{j - 1}}$

Cancel the common factors.

Multiply by $1$ .

$r = \frac{{(- 2)}^{j - 1} {(- 2)}^{j + 0 - (j - 1)}}{{(- 2)}^{j - 1} \cdot 1}$

Cancel the common factor.

$r = \frac{{(- 2)}^{j - 1} {(- 2)}^{j + 0 - (j - 1)}}{{(- 2)}^{j - 1} \cdot 1}$

Rewrite the expression.

$r = \frac{{(- 2)}^{j + 0 - (j - 1)}}{1}$

Divide ${(- 2)}^{j + 0 - (j - 1)}$ by $1$ .

$r = {(- 2)}^{j + 0 - (j - 1)}$

Add $j$ and $0$ .

$r = {(- 2)}^{j - (j - 1)}$

Simplify each term.

Apply the distributive property.

$r = {(- 2)}^{j - j - - 1}$

Multiply $- 1$ by $- 1$ .

$r = {(- 2)}^{j - j + 1}$

Subtract $j$ from $j$ .

$r = {(- 2)}^{0 + 1}$

Add $0$ and $1$ .

$r = {(- 2)}^{1}$

Evaluate the exponent.

$r = - 2$

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

Substitute $1$ for $j$ into $6 {(- 2)}^{j - 1}$ .

$a = 6 {(- 2)}^{1 - 1}$

Simplify.

Subtract $1$ from $1$ .

$a = 6 {(- 2)}^{0}$

Anything raised to $0$ is $1$ .

$a = 6 \cdot 1$

Multiply $6$ by $1$ .

$a = 6$

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

$6 \frac{1 - {(- 2)}^{4}}{1 - - 2}$

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $4$ .

$6 \frac{1 - 1 \cdot 16}{1 - - 2}$

Multiply $- 1$ by $16$ .

$6 \frac{1 - 16}{1 - - 2}$

Subtract $16$ from $1$ .

$6 \frac{- 15}{1 - - 2}$

Simplify the denominator.

Multiply $- 1$ by $- 2$ .

$6 \frac{- 15}{1 + 2}$

Add $1$ and $2$ .

$6 (\frac{- 15}{3})$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$3 (2) \frac{- 15}{3}$

Cancel the common factor.

$3 \cdot 2 \frac{- 15}{3}$

Rewrite the expression.

$2 \cdot - 15$

Multiply $2$ by $- 15$ .

$- 30$

Evaluate the Summation sum from j=1 to 4 of 6(-2)^(j-1)

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