problem:
You have been told that the average rate of inflation between 1910 and 2016 is $3.8\%$. Using this information, calculate how much $100 in 1910 would be worth today.
What I tried:
This one is a little tough because I know inflation is compounded. I know of this formula:
$$ inflation = \frac {final - initial}{initial}$$
Except this formula doesn't consider the fact that inflation would compound upon itself. Do I need to modify the above formula or is there another I can use?
You wish to compute the number of 2016 dollars that are equivalent to $\$100$ 1910 dollars given an average inflation rate of $3.8\%$ per year.
For our purposes, it suffices to treat the inflation rate as constant. Then if $v_i$ is the amount at the beginning of the year, $v_f$ is the amount at the end of the year, and $100r\%$ is the inflation rate, your formula yields $$100r\% = \frac{v_f - v_i}{v_i} \cdot 100\%$$ Solving for $v_f$ yields \begin{align*} r & = \frac{v_f - v_i}{v_i}\\ rv_i & = v_f - v_i\\ rv_i + v_i & = v_f\\ v_i(1 + r) & = v_f \end{align*} Let $v_0$ be the initial amount. Let $v(t)$ be the amount after $t$ years. The calculation we did above shows that $$v(1) = v_0(1 + r)$$ Iterating, we find that \begin{align*} v(2) & = v(1)(1 + r)\\ & = [v_0(1 + r)](1 + r)\\ & = v_0(1 + r)^2\\ v(3) & = v(2)(1 + r)\\ & = [v_0(1 + r)^2](1 + r)\\ & = v_0(1 + r)^3\\ v(4) & = v(3)(1 + r)\\ & = [v_0(1 + r)^3](1 + r)\\ & = v_0(1 + r)^4 \end{align*} which suggests that $$v(t) = v_0(1 + r)^t$$ which can be proved by mathematical induction.
In your example, $v_0 = \$100$, $r = \frac{3.8\%}{100\%} = 0.038$, and $t = 2016 - 1910 = 116$.