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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?

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  • $\begingroup$ x = 980/(120 * 22)... i tried this but im pretty sure im doing it wrong $\endgroup$ – crazywacko Oct 21 '14 at 15:32
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    $\begingroup$ Hint: The first deposit earns interest for 22 years. The second deposit earns interest for 21 years. The third deposit earns interest for 20 years. Etc. The last deposit, made at the beginning of the 20th year, earns interest for 3 years. Can you find an expression for the total simple interest earned by all of these deposits? $\endgroup$ – MPW Oct 21 '14 at 15:50
  • $\begingroup$ something like this 980 = 120(1+ ((x/1)))^22 ? @MPW $\endgroup$ – crazywacko Oct 21 '14 at 16:11
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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 $$120\cdot\left(\underbrace{(1.032\bar6)^{3} + (1.032\bar6)^{4} + \cdots +(1.032\bar6)^{22}}_{20\textrm{ terms}} - (\underbrace{1+1+\cdots+1}_{20\textrm{ terms}}) \right)$$ $$ = 120\cdot\left((1.032\bar6)^3\left(\frac{(1.032\bar6)^{20} - 1}{(1.032\bar6) - 1}\right) - 20\right)$$ $$=\boxed{\boxed{1248.78}} $$

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