Question

(a) Suppose you wish to use the Principle of Mathematical Induction to prove that n(n+1) 1+ 2+ ... +n= - for any positive integer n. i) Write P(1). Write P(6. Write P(k) for any positive integer k. Write P(k+1) for any positive integer k. Use the Principle of Mathematical Induction to prove that P(n) is true for all positive integer n. (b) Suppose that function f is defined recursively by f(0) = 3 f(n+1)=2f (n)+3 Find f(1), f (2), f (3) and f(4).

Answer 1

d) Suppose you wish to use the principle of Mathematical n(n+1) 1+2+ ..... + = - foo any Induction to prove that positive Integer ni i Write PCI). when 1:1 PU - 1KA 2. 1(1+1) P(1) - 1 (ii) Woite P(6) When 0-6 PC6) 1+ 2+ 3+4+5+6 &(1) 21 6(6+1) - P(6) - 21 P(K). foo Trio Write any positive intege% K. K(R - ) P(K) = 1+ 2+ 3 + ... + (iv) Write PC Kit 1) for any positive integer k. K+1(K+ (ku) (K+1)+1) PCK+1) 1+ 2+ 3+...+k+k+1 :- 2 : P(k+1) (kºrv (k+9)

Mathematical the to Induction prove (v) Use that principle of is true Pon) to all positive integer n. k+U(K+1+1) 1 2 +3+4 + .. +K+K+ PCK) KCK : Equate Substitute pik) in Pikl 2 PCK+1) place of 1+2+...+k !) + K+1 = (*+ 1) (k+2) KCK+ 1) + 2(x+1) = (K+U) [k+2) Denominatoos same, so KCK+1) + 30K+1) = (K+1)(K+2) K?+k+ 2K +2 = K2 + 2K+ k +2 K? +3K+2 K?+ 3K+2 .. P(n) is pim) is toue toue for for PLK) & P(ktl). all positive Integers of n.
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is that function defined recursively by b) Suppose f(0) =3 #(m+1) : 2 (0) +3 find f(0), 4(2), f(3) at F14). sol! when ne 0 f(0+1) = 27(0) + 3 2 ( 3) + 3 ' f(1) = f(0) = 3 =>(given) 6(1) 6+3 -9 when : +1 + 1) = 2f( 1) + 3 f(2) : 27(1) +3 . f(t): 9 +(23 = 2(9)+3 :5712) = 21 When n: 2 +(2+1) 26(2) + 3 f(3) = 2(21) +3 :{f(3) = 45 : f[2) = 21 When n=3 +(3+1) = 2#(3) + 3 +14) = 2(45) +3 : f(3) = 4 5 If (4) = 93