Suppose you have a bank account with a € 100.- initial deposit which once a year is compounded with \(\gamma = 3\% \) interest. In this case your gain is \(0.03*100 = 3\), which means that according to the equation

\[S_{t+1}=S_t+gain\]your bank account will hold € 103.- after the first year.

\[103 = 100 + 0.03 * 100\]After the second year your \(gain\) again will be \(\gamma*S_t\) with \(\gamma\) still being 0.03 but \(S_t\) now grown to 103. Hence

\[S_{t+2}= 103 + 0.03 * 103 = 106.09\]After the third year your bank account will hold

\[S_{t+3}=106.09 + 0.03 * 106.09 = 109.2727\]And in general your account will grow according to the difference equation.

\[S_{t+1}=S_t + \gamma * S_t \text{ respectively } S_{t+1}=S_t * (1+ \gamma)\]With constant \(\gamma\) hence, starting from an initial deposit of \(S_0\), your account will grow in t time steps according to

\[S_t=S_0+\gamma*S_0+\gamma*S_1+\gamma*S_2+...+\gamma*S_t\]or short

\[S_t=S_0*(1+\gamma)^t\]According to this formula, after 10 years of annually compounding 3% interest your bank account should hold (rounded to two decimal places):

\[S_{10}=S_0*(1+\gamma)^{10}=100*(1+0.03)^{10} = 134.39 \text{Euro}\]