To save for Malcom's education, his parents put $1,200 into a savings account when
Malcom was born in 2006. Every birthday, Malcolm's parents deposit 10% more than the
previous year.
Part A
In this problem we have an exponential function of the form
y=a(1+r)^x
where
a is the initial deposit
r is the rate of change
x is the number of years
y is the amount deposited in the saving account
so
we have
a=$1,200
r=10%=0.10
substitute
y=1,200(1+0.10)^x
y=1,200(1.10)^x --------> explicit formula
Part B
2006 -----> For x=0 -------> y=$1,200
2010 -----> For x=4 -----> y=$1,756.92
2014 ----> For x=8 -----> y=$2,572.31
2018 ----> For x=12 ----> y=$3,766.11
2022 ----> For x=16 ----> y=$5,513.97
Part C
we have an geometric sequence
To find out the sum
we use the formula
![S=(a(1-r^n))/(1-r)](https://img.qammunity.org/2023/formulas/mathematics/college/3exeaw3nhm6ns5ng5yzxiks5ypvh41dct1.png)
where
a is the first term
n is the number of terms
r is the common ratio
In this problem
a=$1,200
r=1.1
n=18
substitute
![\begin{gathered} S=(1,200(1-1.1^((18))))/(1-1.1) \\ S=\$54,719.01 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vsble3onsrfttiywqa3vkm432z9kag1apv.png)
answer Part C is $54,719.01