Given the recurrence equation `T(n) = T(n-2) + 2n` for `n >= 2` with base cases `T(0) = 1` and `T(1) = 3`.We can solve the given recurrence using recursion tree and backward substitution method.
To solve the recurrence equation `T(n) = T(n-2) + 2n`, for the given base cases we have,T(0) = 1, and T(1) = 3.Using recurrence relation, `T(n) = T(n-2) + 2n`, we can get the following values of T(n):`T(2) = T(0) + 2*2 = 1 + 4 = 5` `T(3) = T(1) + 2*3 = 3 + 6 = 9` `T(4) = T(2) + 2*4 = 5 + 8 = 13` `T(5) = T(3) + 2*5 = 9 + 10 = 19` `T(6) = T(4) + 2*6 = 13 + 12 = 25` `T(7) = T(5) + 2*7 = 19 + 14 = 33` `T(8) = T(6) + 2*8 = 25 + 16 = 41` `T(9) = T(7) + 2*9 = 33 + 18 = 51` `T(10) = T(8) + 2*10 = 41 + 20 = 61` `T(11) = T(9) + 2*11 = 51 + 22 = 73` `T(12) = T(10) + 2*12 = 61 + 24 = 85` `T(13) = T(11) + 2*13 = 73 + 26 = 99` `T(14) = T(12) + 2*14 = 85 + 28 = 113` `T(15) = T(13) + 2*15 = 99 + 30 = 129` `T(16) = T(14) + 2*16 = 113 + 32 = 145`Therefore, the values of `T(n)` for `n >= 2` is given by `T(n) = 1, 3, 5, 9, 13, 19, 25, 33, 41, 51, 61, 73, 85, 99, 113, 129, 145, 163, 181, ...`.Hence, the solution to the given recurrence equation is `T(n) = 2*[(n/2)^2] + 1` for `n >= 0` where `[.]` represents the greatest integer function. Therefore, the above answer satisfies the given recurrence equation.