How would I solve this recurrence relation without using the Master's Theorem?
T(n) = T(n − 2) + 5n; T(1) = T(2) = 1
Solve recurrence relation without master theorem
2
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recurrence-relations
recursion
1 Answers
1
- For even $n$:
$T(n) = 5n + 5(n-2) + 5(n-4) + ... + 5(n-(n-2)) = 5\sum_{i=0}^{(n/2)-1}(n-2i) = 5n(n/2) - 10\sum_{i=0}^{(n/2) -1}i = 5n(n/2) - 10(\frac{\big((n/2) -1\big)(n/2)}{2})$
- For odd $n$:
$T(n) = 5n + 5(n-2) + 5(n-4) + ... + 5(n-(n-1)) = 5\sum_{i=0}^{(n-1)/2}(n-2i) = 5n(\frac{n-1}{2}+1) - 10\sum_{i=0}^{(n-1)/2}i = 5n(\frac{n-1}{2}+1) - 10(\frac{\frac{n-1}{2}(\frac{n-1}{2}+1)}{2})$