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 can write this:
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:
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.