problem: Compute $i+i^2+i^3+\cdots+i^{258}+i^{259}$.
solution: Each group of 4 consecutive powers of $i$ adds to 0: \[ i + i^2 + i^3 + i^4 = i - 1 - i +1 = 0,\] \[ i^5+i^6+i^7+i^8 = i^4(i+i^2+i^3+i^4) = 1(0) = 0, \] and so on.  Because $259 =64\cdot4+3$, we know that if we start grouping the powers of $i$ as suggested by our first two groups above, we will have 64 groups of 4 and 3 terms left without a group: $i^{257}+i^{258}+i^{259}$. To evaluate the sum of these three terms, we use the fact that $i^{256}=(i^4)^{64}=1^{64}$, so \[ i^{257}+i^{258}+i^{259}=i^{256}(i+i^2+i^3)=1(i-1-i)=-1. \] So \begin{align*}
  &\quad i+i^2+i^3+\cdots+i^{258}+i^{259} \\
  &= (i+i^2+i^3+i^4) + (i^5+i^6+i^7+i^8) + \cdots \\
  &\quad + (i^{253}+i^{254}+i^{255}+i^{256}) + (i^{257}+i^{258}+i^{259}) \\
  &= 0 + 0 + \cdots + 0 + -1 \\
  &= \boxed{-1}.
  \end{align*}