First I want to say thank you for the detailed explanation of the exercise that really helped me to wrap my head around the problem. I found though a little issue with the formula $$T(n,2) = \frac{n^2-3n}5+1$$. When n=12 it results in 22.6 which is more than twice as less the correct answer 49(as seen on the image). Although math is not of my strengths I suspected that starting the summation with index of zero is somehow wrong and I ended up with the following formula:
$$T(n,2) = Ceil\left(\frac n5\right)+1+\sum_{i=1}^{Ceil(n/5)}T(n-5(i-1),2)$$
From this formula I derive two, because of the ceiling. When the amount is divisible by 5 without reminder:
$$T(n,2) = \left(\frac n5\right)+1+\sum_{i=1}^{n/5}T(n-5(i-1),2)$$
$$T(n,2) = \frac{n^2+7n}5+1$$
With reminder:
$$T(n,2) = \left(\frac n5+1\right)+1+\sum_{i=1}^{(n/5)+1}T(n-5(i-1),2)$$
$$T(n,2) = \frac{n^2+7n}5+3$$
The latter one gives as result 48,6 $$\approx$$ 49 when the amount is 12. The former one gives correctly 35 when the amount is 10. I hope that is helpful. Thank you again for sharing this valuable knowledge.
First I want to say thank you for the detailed explanation of the exercise that really helped me to wrap my head around the problem. I found though a little issue with the formula$$T(n,2) = \frac{n^2-3n}5+1$$ . When n=12 it results in 22.6 which is more than twice as less the correct answer 49(as seen on the image). Although math is not of my strengths I suspected that starting the summation with index of zero is somehow wrong and I ended up with the following formula:
From this formula I derive two, because of the ceiling. When the amount is divisible by 5 without reminder:
With reminder:
The latter one gives as result 48,6$$\approx$$ 49 when the amount is 12. The former one gives correctly 35 when the amount is 10. I hope that is helpful. Thank you again for sharing this valuable knowledge.