Find an explicit formula for the geometric sequence \dfrac12\,,-4\,,\,32\,,-256,.. 2 1 ​ ,−4,32,−256,..start fraction, 1, divided by, 2, end fraction, comma, minus, 4, comma, 32, comma, minus, 256, comma, point, point. Note: the first term should be \textit{a(1)}a(1)start text, a, left parenthesis, 1, right parenthesis, end text. a(n)=a(n)=a, left parenthesis, n, right parenthesis, equals

using arithmetic sequence formula

Step-by-step explanation:

and geometric sequence to solve the following questions

a(n)= 1/2 * (-8) n-1

Step-by-step explanation:

In a geometric sequence, the ratio between successive terms is constant. This means that we can move from any term to the next one by multiplying by a constant value. Let's calculate this ratio over the first few terms:

\dfrac{-256}{32}=\dfrac{32}{-4}=\dfrac{-4}{\frac12}=\blue{-8}  

32

−256

​

=  

−4

32

​

=  

2

1

​

−4

​

=−8start fraction, minus, 256, divided by, 32, end fraction, equals, start fraction, 32, divided by, minus, 4, end fraction, equals, start fraction, minus, 4, divided by, start fraction, 1, divided by, 2, end fraction, end fraction, equals, start color #6495ed, minus, 8, end color #6495ed

We see that the constant ratio between successive terms is \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed. In other words, we can find any term by starting with the first term and multiplying by \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed repeatedly until we get to the desired term.

Let's look at the first few terms expressed as products:

nn111222333444

h(n)\!\!\!\!\!h(n)h, left parenthesis, n, right parenthesis\red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large0}\!\!\!\!\!\!  

2

1

​

⋅(−8)  

0

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, start superscript, 0, end superscript\red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large1}\!\!\!\!\!\!  

2

1

​

⋅(−8)  

1

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, start superscript, 1, end superscript\red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large2}\!\!\!\!\!\!  

2

1

​

⋅(−8)  

2

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, squared\red{\dfrac12}\cdot\!\!\!\left(\blue{-8}\right)^{\,\large3}  

2

1

​

⋅(−8)  

3

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, cubed

We can see that every term is the product of the first term, \red{\dfrac12}  

2

1

​

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, and a power of the constant ratio, \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed. Note that this power is always one less than the term number nnn. This is because the first term is the product of itself and plainly 111, which is like taking the constant ratio to the zeroth power.

Thus, we arrive at the following explicit formula (Note that \red{\dfrac12}  

2

1

​

start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030 is the first term and \blue{-8}−8start color #6495ed, minus, 8, end color #6495ed is the constant ratio):

a(n)=\red{\dfrac12}\cdot\left(\blue{-8}\right)^{\large{\,n-1}}a(n)=  

2

1

​

⋅(−8)  

n−1

a, left parenthesis, n, right parenthesis, equals, start color #df0030, start fraction, 1, divided by, 2, end fraction, end color #df0030, dot, left parenthesis, start color #6495ed, minus, 8, end color #6495ed, right parenthesis, start superscript, n, minus, 1, end superscript

Note that this solution strategy results in this formula; however, an equally correct solution can be written in other equivalent forms as well.

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.