Question
Each year, Cathy invests $1,200 in
her account. The account pays an
interest rate of 6.3%. The formula to
calculate the balance in her account
is B =A(1+ r)n+1- A, where
r
A is the amount invested per year,
r is the interest rate, and n is the
number of years investing
her account. The account pays an
interest rate of 6.3%. The formula to
calculate the balance in her account
is B =A(1+ r)n+1- A, where
r
A is the amount invested per year,
r is the interest rate, and n is the
number of years investing
Answers
MathMate
Not sure what your question is, but the formula does not look right. You may want to check the post for typos.
If someone invests $A at the beginning of the year, at the end of one year, the balance is $A plus the interest, which is $Ar. So the balance is A(1+r).
For two years, it would be the sum of two investments,
A(1+r)²+A(1+r)
for n years, it would then be:
A[(1+r)+(1+r)²+(1+r)³...+(1+r)^n]
which is algebraically equal to
A[(1+r)^(n+1)-1)]/r - A
=A[((1+r)^(n+1)-1)/r - 1]
slightly different from yours. Your formula seems to miss out the "/r".
If someone invests $A at the beginning of the year, at the end of one year, the balance is $A plus the interest, which is $Ar. So the balance is A(1+r).
For two years, it would be the sum of two investments,
A(1+r)²+A(1+r)
for n years, it would then be:
A[(1+r)+(1+r)²+(1+r)³...+(1+r)^n]
which is algebraically equal to
A[(1+r)^(n+1)-1)]/r - A
=A[((1+r)^(n+1)-1)/r - 1]
slightly different from yours. Your formula seems to miss out the "/r".