Question

Find the exact solution to the following recursive formula

a. T(1) = 1 and T(n)= T(n-1) + 3. Note the n-1 means that powers of two aren't the feature to use here. b. T(1) = 1 and T(n) = T(n-1) + (2n-1) c. T(1) = 1 and t(n) = T(n/2) + Vn (master theorem is useful here)

Answer 1

The recursion formula for a) and b) is shown below. For c) master theorem is used as asked in c) part.

a) 7(1)-1 T(n)= Tari) +3. T(2) t = 1 (1)+3 = 1+ 3 = 1+ 3x) T(3) = T (2) + 3 = 1+3+3 = 1+ 3x2 T14) = T13) + 3 = 1+3+3+3=1+3+3 Tln) Tlui) + 3= 1+ 3 (n-1) Hence, T(n) = 3n-2

b) Tln)= T(n-1) + 2n-1 T(1) = 1 T (2) = T11) + 3 to =1+3=4 4 T(37= T (2)+5= 9 T(4) = T(3) + 7 = 16 . Ting = n2. We can also see it as T (n)- Tut) 2n-1 n² Caril? zurl W W + 2n-1 curl 2n-12 2url tena, T(n) = n2

c) 7(1)=1 By T(n) = T(1/2) + un Master theorem, we get Tula at(n/ble fla) a=1, 6:2, flut vu It satisfies the condition f(nl= recalog, at e) where & so& constant Sn = RC log, 1+E) уи when {= /, it is true. tence, from master theorem, T(nla o crn) (mt)