Question

Solve the following recurrences using substitution. (n)T(n 2)3n + 4,for all n 2 3. G iven T(1) = 1, and T(2) 6

Answer 1

(T/M-2) + 3n44 : m=2 n33 T(n) = T (m2) + on the T(n-2) = T (m-4) + 3(n-1) + 4 T(n-4) = T (16) + 3 (-4) +4 substituting T(m) = the above values in Tlm) T(m-4) + 3 (0-2) +4 + 3nt4 = T(m-t) + 3(-2) + + 3 (-2) +4 + 3 m+ - T(n-6) + 3nx3 - 12-6 4 4x3. TM-6) + 3x3 - 18 + 4x3 . T(m)_= T(m-2k) + 3m»k - pk x3 +hxk I) Giver , T EL. bo n -2k = 1 -> k-n-1 from 0, - T(m) T(1) + 3m x n 1 - 2 - 1) + 3 + 4x (1-1) 0(1) + 0(m2) - Olm) + O(n) O(ne) T(m) =

I Given So, if T(m) T (2) = 6 nok = 2 → k = n-2 from im a T (2) + 3nx T (2) + 3nx n-2 - 2 x 1-2 & 3 + 4x n. 2 + 0/n") - 0(m) + 0(m) 2 0(1) T(m) 2 0(m²)