deletions | additions       diff --git a/a) Number of blits.tex b/a) Number of blits.tex  index d914a3f..f600f99 100644  --- a/a) Number of blits.tex  +++ b/a) Number of blits.tex 
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\begin{subequations}  \begin{align}  T(2^1) &= 5\\  T(2^2) &= 5 + 5 * \cdot  2^2\\ T(2^3) &= 5 + 5 * \cdot  2^2 + 5 * \cdot  2^4\\ T(2^4) &= 5 + 5 * \cdot  2^2 + 5 * \cdot  2^4 + 5 * \cdot  2^6\\ T(2^5) &= 5 + 5 * \cdot  2^2 + 5 * \cdot  2^4 + 5 * \cdot  2^6 + 5 * \cdot  2^8 \end{align}  \end{subequations} 
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T(2^n) &= \sum_{i = 0}^{n-1} \left[5 \cdot 4^i\right]  = 5\sum_{i = 0}^{n-1} \left[4^i\right]\\  &= 5\frac{1 - 4^n}{1 - 4}  = \frac{5 - 5 * \cdot  4^n}{-3}\\ &= \frac{5 * \cdot  4^n - 5}{3} \end{split}  \end{equation}  Then:  \begin{equation}  \begin{split}  T(2^{n+1}) &= 5 + 4\frac{5 * \cdot  4^n - 5}{3}\\ &= 5 + \frac{5 * \cdot  4^{n+1} - 5*4}{3}\\ 5 \cdot 4}{3}\\  &= \frac{5 * \cdot  4^{n+1} + 5 * 3 - 5 * \cdot  4}{3}\\ &= \frac{5 * \cdot  4^{n+1} - 5}{3} \end{split}  \end{equation}  Which gives us that: