Suppose your parents deposited $1500 in an account paying 3.5% interest compunded annually(once a year) when you ere born. Find the account balance after 18 years.
So how do you plug all this information into this formula: B=p(1+r)^x?
What will an amount of R dollars deposited in a savings bank at the end of each year for N years, and earning I% compounded annually, amount to in N years.
Stated another way, what will R dollars deposited in a bank account at the end of each of n periods per year, and earning interest at I/n, compounded n times per year, amount to in N years?
Remember, an ordinary annuity consists of a definite number of deposits made at the ENDS of equal intervals of time. An annuity due consists of a definite number of deposits made at the BEGINNING of equal intervals of time.
For an ordinary annuity over n payment periods, n deposits are made at the end of each period but interest is paid only on (n - 1) of the payments, the last deposit drawing no interest, obviously. In the annuity due, over the same n periods, interest accrues on all n payments but there is no payment made at the end of the nth period. Consider the deposit of $R at the end of each year for 5 years at an iterest rate of I% compounded annually.
The compound interest on the first deposit at the end of the first year is S1 = R(1 + i)^(n-1).
The compound interest on the second deposit at the end of the second year is S2 = R(1+i)^(n-2).
The compoud interest on the third deposit at the end of the third year is S3 = R(1+i)^(n-3)
The compound interest on the fourth deposit at the end of the fourth year is S4 = R(1+i).
There is no interest gained on the fifth deposit at the end of the fifth year.
The final accumulation is S = R(1+i)^(n-1) + R(1+i)^(n-2) + R(1+i)^(n-3) + R(1+i) + R.
Reversing this, S = R + R(1+i) + R(1+i)^(n-3) + R(1+i)^(n-2) + R(1+i)^(n-1)
Note that this is a geometric progression with first term a = R, common factor r = (1+i) and the number of terms n = n.
As we know, the sum of the terms of such a progression derives from S = a(r^n - 1)/((r - 1).
Therefore, the sum total of our comounded amounts becomes S = R[(1+i)^n - 1]/(1+i-1) = R[(1+i)^n - 1]/i.
The formula for determining the accumulation of a series of periodic deposits, made at the end of each period, over a given time span, an ordinary annuity, is
S(n) = R[(1 + i)^n - 1]/i
where S(n) = the accumulation over the period of n intervals, R = the periodic deposit, n = the number of interest paying periods, and i = the annual interest % divided by 100 divided by the number of interest paying periods per year.
When an annuity is cumputed on the basis of the payments being made at the beginning of each period, an annuity due, the total accumulation is based on one more period minus the last payment. Thus, the total accumulation becomes
S(n+1) = R[(1 + i)^(n+1) - 1]/i - R = R[[{(1 + i)^(n + 1) - 1}/i] - 1] or S
S(n+1) = R[(1 + i)^(n + 1) - (1 + i)]/i
Example: $200 deposited annually for 5 years at 12% annual interest compounded annually. Therefore, R = 200, n = 5, and i = .12.
Ordinary Annuity
S = R[(1 + i)^n - 1]/i = 200[(1.12)^5 - 1]/.12 = $1270.56
Annuity Due
Sn = [R[(1 + i)^(n +1) - 1]/i - R] = 200[(1.12)^6 - 1]/.12 - 200 = $1,423.03
I'll let you plug in your numbers.
Suppose your parents deposited $1500 in an account paying 3.5% interest compunded annually(once a year) when you ere born. Find the account balance after 18 years.
So how do you plug all this information into this formula: B=p(1+r)^x?
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