Find the Sum of the First 6 Terms 2 , 8 , 32 , 128

$2$ , $8$ , $32$ , $128$

Step 1

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by $4$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = 4$

Step 2

This is the form of a geometric sequence.

$a_{n} = a_{1} r^{n - 1}$

Step 3

Substitute in the values of $a_{1} = 2$ and $r = 4$ .

$a_{n} = 2 \cdot 4^{n - 1}$

Step 4

Multiply $2 \cdot 4^{n - 1}$ .

Rewrite $4$ as $2^{2}$ .

$a_{n} = 2 \cdot {(2^{2})}^{n - 1}$

Multiply the exponents in ${(2^{2})}^{n - 1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a_{n} = 2 \cdot 2^{2 (n - 1)}$

Apply the distributive property.

$a_{n} = 2 \cdot 2^{2 n + 2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$a_{n} = 2 \cdot 2^{2 n - 2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$a_{n} = 2^{1 + 2 n - 2}$

Subtract $2$ from $1$ .

$a_{n} = 2^{2 n - 1}$

Step 5

This is the formula to find the sum of the first $n$ terms of the geometric sequence. To evaluate it, find the values of $r$ and $a_{1}$ .

$S_{n} = \frac{a_{1} (r^{n} - 1)}{r - 1}$

Step 6

Replace the variables with the known values to find $S_{6}$ .

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

Step 7

Simplify the numerator.

Raise $4$ to the power of $6$ .

$S_{6} = 2 \cdot \frac{4096 - 1}{4 - 1}$

Subtract $1$ from $4096$ .

$S_{6} = 2 \cdot \frac{4095}{4 - 1}$

Step 8

Simplify the expression.

Subtract $1$ from $4$ .

$S_{6} = 2 \cdot \frac{4095}{3}$

Divide $4095$ by $3$ .

$S_{6} = 2 \cdot 1365$

Multiply $2$ by $1365$ .

$S_{6} = 2730$

Step 9

Convert the fraction to a decimal.

$S_{6} = 2730$

Find the Sum of the First 6 Terms 2 , 8 , 32 , 128

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