need help calculating the interest "i"

Question

A regular deposit of 120 dollar made at the beginning of each year for 20 years. Simple interest is Calculated at a rate of i per year for 22 years. At the end of the 22-year period, the total interest in the account is $980. Suppose that interest at the rate i compounded annually had been paid instead. How much interest would have been in the account at the end of the 22 years?

Answer 1

The amount of simple interest earned by investing $P$ dollars for $t$ years at an annual simple interest rate of $i$ is $P\cdot i \cdot t$.

So $120$ invested at the beginning of the first year earns $\boxed{120\cdot i \cdot 22}$, since it accumulates for 22 years.

The $120$ invested at the beginning of the second year accumulates for one less year (you waited a year before making that deposit), so it earns $\boxed{120\cdot i\cdot 21}$ in simple interest by the end of the period.

Similarly, the $120$ invested at the beginning of the third year accumulates for only 20 years, so it earns $\boxed{120\cdot i\cdot 20}$ by the end of the period.

And so on, all the way down to the last deposit of $120$, invested at the beginning of the twentieth year; it earns $\boxed{120\cdot i \cdot 3}$ by the end of the period.

Adding up all of the interest, we have that $$980 = 120\cdot i \cdot(3 + 4 + 5 + \cdots + 21 + 22)$$ $$980 = 120\cdot i \cdot 250$$ $$i = \frac{980}{120\cdot 250} = 0.032\bar6 = \boxed{\boxed{3.2\bar6\%}}$$

Now compute what the corresponding compound interest amounts would have been. The relevant formula here is that $P$ dollars compounded annually at a rate $i$ for $t$ years grows to $P\cdot (1+i)^t$, so the amount of interest earned is $P\cdot ((1+i)^t - 1)$.

The $120$ invested at the beginning of the first year earns $\boxed{120\cdot((1.032\bar6)^{22}-1)}$ in compound interest by the end of the period.

The $120$ invested at the beginning of the second year earns $\boxed{120\cdot((1.032\bar6)^{21}-1)}$ in compound interest by the end of the period.

The $120$ invested at the beginning of the third year earns $\boxed{120\cdot((1.032\bar6)^{20}-1)}$ in compound interest by the end of the period.

And so on, all the way down to the last deposit of $120$, which earns $\boxed{120\cdot((1.032\bar6)^{3}-1)}$ in compound interest by the end of the period.

The total compound interest earned is $$120\cdot\left(\underbrace{(1.032\bar6)^{3} + (1.032\bar6)^{4} + \cdots +(1.032\bar6)^{22}}_{20\textrm{ terms}} - (\underbrace{1+1+\cdots+1}_{20\textrm{ terms}}) \right)$$ $$ = 120\cdot\left((1.032\bar6)^3\left(\frac{(1.032\bar6)^{20} - 1}{(1.032\bar6) - 1}\right) - 20\right)$$ $$=\boxed{\boxed{1248.78}} $$