Question

2 (25 pts). Let an algorithm has complexity S(n)=S(n-1)+f(n), where for k=1,2,3,... f(k)=k+k/3. Answer these two questions: (1) Find the closed form for S(n) if S(2)=1. (2) Prove by mathematical induction that the closed form you found is correct.

Answer 1

Here given that - sini= sint) + find, 1 where for K=1, 2, 3, ... flk) = x2 + K/3. a). if SC2) = 1. for equation (1) sing-sentt fin) = S(n)-sent) = fem) - () putting nal, S(1) - S10) -() putting naz $(2) - Sl) = tez) Adding these two equation, we have S(22-510) febffP2) :5(2) II then = 5(2) -S(O) = (2+1)+(4+2) By defn of feks = 1 -S(O) = 2 + 1 + 4 + 2 -S(O)25 => 15102-5 - 1

Page : Date: From equation (1), we can write, $(1) - S(0) - FL) $(2) - S(1) ? H2) $(3) - S02) - H3) 384) -503) 2 714) sen2 - Sent) - Hen) Adding these set of equations, we have srn) scol= fel) + f2) t-atten) = sen) - 5(0) = (22 +) +(2*4 23 }--- (24 /s) =) Sin) -5102 = 3 (1+2+ -- + n) + (12+22+---+n?) → Sem) – 5(0) n (intl) + (141) (2n1) = S(N) -5(0) m(nal + min ) (21 41 ) 5 sem -S(O) = n(nt (14 20:41] 7 scho-sco) = n(n+1)(n+1) 3 - she nintl)2+510) Sana bintia-s = $ [ n(11)248]

(2). Here we found, sm) = {[ n(941) +-15]. tij for nal SCP) = { [2(141)2-15) - . Also a s10)+ f(1)2-54.1752 1543+1 --44 Flaws, we get 2 SIL) = -. [5(0)+(13]. ie statement true for nas... - let assume that statement sing is true for nam. ie . - Scm = ₃ [m (m+2)2-15]. .- -tij statement is also Then we true show for that namtl. : Form (ii) s(m) = {[m(m +1) ?-15]. 3 Adding /(m + 1 ) 2 + (m+1) y both sides of the above equation we have sim) tf (m+1)2.4 (= (m (+1) ?-15]+ (m tijet (mm) }

Date: / / = s(m) + f(m + 1) = m (m + 1)2 - 5 + (n + 1)2 + (m+1). = s[mt) = (m++) momt! +3(m+1) +37-5 3 l (+ 1) = ( 1 ) ( x + m + 3m + 3 + 7- mt! [ m2 4 4m +47-5 = (met (m+212_ mtl I m² + 4mt 3 L - 3 simti) = § f (m+1) ? [( m+)+1]?-15] scn) is sue from equation (ili), we have na mtl. By principle of induction, Thus stallment scor is true for tnEN. I sens = 1/3 [ m (2412²-15]