On Gideon’s first birthday, Dr Prince deposits \$1400. 4.6% as a decimal is 0.046, so on the second birthday, the original \$1400 grows by the factor 1.046, becoming \$1464.40. Dr Prince deposits a further \$1400, making the amount at the start of the second year \$2864.40.

On the third birthday \$2864.40 will grow by the factor 1.046 to \$2996.16.

Let’s write this down symbolically. Let the deposit be D and the rate r (so D=1400 and r=0.046).

Growth at end of 1st year=D(1+r)

Amount at beginning of 2nd year=D+D(1+r)

Growth at end of 2nd year=(D+D(1+r))(1+r)=D(1+r)+D(1+r)²

Amount at beginning of 3rd year=D+D(1+r)+D(1+r)²

Growth at end of 3rd year=(D+D(1+r)+D(1+r)²)(1+r).

So we can write this as D(1+r+(1+r)²+(1+r)³).

Fast forward to the end of the nth year, Gideon’s (n+1)th birthday, assuming that his first birthday is when he is 1 year old.

We have:

D(1+r+(1+r)²+(1+r)³+...+(1+r)ⁿ).

We can write this:

D(1+r)[1+1+r+(1+r)²+(1+r)³+...+(1+r)ⁿ⁻¹].

In the square parentheses we have a geometric progression so we can use the sum formula: S(n)=[(1+r)ⁿ-1]/[1+r-1]=[(1+r)ⁿ-1]/r.

At the end of the nth year the accumulated amount is:

D(1+r)S(n)=[D(1+r)/r][(1+r)ⁿ-1].

Let’s check this by using the given values to discover what has accumulated by the time Gideon is 3 years old, so n+1=3, n=2:

(1400×1.046/0.046)(0.094116)=\$2996.16, which is what was calculated earlier.

You have not asked a specific question, but you now have a formula for calculating how much Gideon would have by the time he is, say, 18 years old.

by Top Rated User (761k points)