Selected Exercises From The First Four Chapters of Grimaldi and Raman's Discrete and Combinatorial Mathematics

 

Selected Exercises from Ralph P. Grimaldi and B. V. Raman’s  Discrete and Combinatorial 

Mathematics: Mathem atics: An Applie Applied d Intro Introducti duction  on  October 3, 2011 1.1-2.4.a. There are ten candidates and four positions, thus the number of possible slates is P  is  P (10 (10,, 4) =   10! 6!   = 10 · 9 · 8 · 7. 1.1-2.4.b.i. 1.1-2. 4.b.i. Let Let   a, b   and and   c   denote denote the three pys pysician icians. s. If   a  is the president, then   9! 6!  possible slates with there remainnt. 3 positions 9 candidates, there preside The sameand amount of slates implying are possible if   bare   or  or   c  are president. Thu president. Thuss a  as president.   9! the total number of slates with a physician as the president is 3 · 6!

1.1-2.4.b.ii. We will count all slates having a having  a and  and not having b having  b or  or  c and  c  and then multiply our answer by three, to include the slates containing exactly one of the other physician phy sicians. s. Notic Noticee that the number of linear arrangem arrangement entss of thre threee non-phys non-physician ician   7! board members is   P (7 P (7,, 3) = 4! . Eac Each h of the these se linea linearr arr arrang angeme ement ntss cor corres respond pond to 4 slates containing  containing   a  and neither   b   nor   c, since   a   can be inserted in any of the four positions. position s. Thu Thuss the total number of slates containi containing ng exactly one phy physicia sician n is 3 · 4 · 7! 4! 7! 1.1-2.4.b.iii. 1.1-2.4.b.ii i. The num number ber of slate slatess con containi taining ng no phy physician sicianss is  is   P (7 P (7,, 4) =   3! . By the rule of the sum, the number of slates with at least one physician is equal to the number of all possible slates minus the number of slates with no physicians, which is 7! equal to   10!   −   3! . 6!

1.1-2.12.  act, atc, tac, tca, cta, cat  1.1-2.24.   P ( P (n + 1, 1, r) =

  (n+1)! n! n+1) n+1) n!   = (n  +1(n−+1)   =  (n(+1   =  (n(+1 P ( P (n, r ) (n+1−r)! r )(n−r )! −r ) (n−r )! −r )

1.1-2.35.a. This is a simple circ 1.1-2.35.a. circular ular arrange arrangemen mentt problem. The num number ber of wa ways ys is 6!

7! 7 =

1.1-2.35.b. First cons 1.1-2.35.b. consider ider thes thesee two people as a single unit. Thu Thuss we count only circular circu lar arrange arrangemen ments ts of 6 ob object jects, s, of whic which h there are 5!. Note that eac each h of these 5! arrangements correspond to 2 arrangements, because the two inseparable friends can 1

 

be arranged in two ways. Thus the total number of arrangements is 2 · 5! 13 . Simila Similarly 1.3.8.a. First 1.3.8.a. noti notice ce that num number ber of wa ways ys to choose five clubs is 5  13 rly, , 13 13 there are 5  ways to choose five diamonds, 5  ways to choose five hearts, and 5 ways wa ys to choose choose five club clubs. s. By the rul rulee of the sum, the tota totall nu numbe mberr of way wayss to get 13   4 13! five cards of the same suit is 4 5  = 5!8!

 

·

1.3.8.b. Four of the five  in the hand are already fixed. The number of ways 52cards 4 to choose the last card is 1  = 48. −

1.3.8.c. As discusse 1.3.8.c. discussed d above, there are 48 way wayss to get four aces. Simil Similarly arly,, there are 48 ways to get four twos, 48 ways to get four threes, etc. By the rule of the sum, the total number of ways to get four of a kind is 13 · 48 = 624. 4 1.3.8.d. The number 4of ways to choose three aces is 3  = 4. The number of ways to choose two jacks is 2  = 6. By the rule of the product, the total number of ways to choose three aces and two jacks is 4 · 6 = 24.



1.3.8.e. From the previous question we know that there are 4 ways to choose three aces. ace s. Jus Justt like like the there re are 6 wa ways ys to ch choose oose tw twoo jac jacks, ks, ther theree are 6 wa ways ys to choose 2 tw twos, os, 6 wa ways ys to choose 2 thre threes, es, 6 wa ways ys to choose 2 fours fours,, etc. The total num number ber of  way to get three aces and a pair is 4 · 12 · 6 = 288. 1.3.8.f. Just lik 1.3.8.f. likee there are 288 way wayss to get 3 aces and a pair, there are 288 ways to get 3 tw twos os and a pai pair, r, 288 way wayss to get 3 three threess and a pai pair, r, etc. Th Thus us the tota totall number of ways to get a full house is 13 · 288 = 3, 3, 744. 4. By 1.3.8.g. The num 1.3.8.g. number ber of wa ways ys to choose three cards of given rank is 43  = 4. the rule of the sum, the number of ways to choose three cards of any one rank is 13 · 4 = 52. There are 48 cards from which to choose the last two cards (since three have been chosen already and one of the unchosen would make the hand into four of    48 a kind). Thus there are 52 2  ways to get a hand containing three cards of the same rank. In order for the hand to be considered three of a kind and not a full house, it must not be any of the 3,744 full houses counte counted 1.3.8.d.. Thu Thuss the total num number ber  d in 1.3.8.d 48 54, 912. of hands which are three of a kind is 52 2 − 3, 744 = 54,



 possibilities for what rank the pairs will be. Furthermore, 1.3.8.h. are 13 2  4There there 4 are 2  possibilities for what the suits of the members of the first pair are, and 2  possibilities for what the suits of the members of the second pair are. There are 44 cards from which to choose the last card (since four cards have already been chosen and four of the unchosen are of the same of the chosen). Thus the total  as two  rank 13 4 4 44   13 12 123, 552 number of way to get two pairs is 2 2 2 1  = 2   6 · 6 · 44 = 123,

 

·

1.3.16.a.

  6

i=1

(i2 + 1) = 2 + 5 + 10 + 17 + 26 + 37 = 97 2

 

1.3.16.b. 1.3.16.c.

  2

 j =−2

  10

i=0

( j 3 − 1) = (− (−2)3 − 1 + (− (−1)3 − 1 + (0)3 − 1 + 12 − 1 + 23 − 1 = 5

[1 + (− (−1)i ] = 2 + 0 + 2 + 0 + 2 + 0 + 2 + 0 + 2 + 0 + 2 = 12

1.3.16.d. 2kn=n (−1)k . Notice that this sequence alternates ones and negative ones and that it starts with a one and ends with minus one, implying that the sum is zero.

 

1.3.16.e.

  6

 i(  i(−1)i = −  −11 + 2 − 3 + 4 − 5 + 6 = 3

i=1

1.3.26.a. 1.3.26 .a. In order to mak makee the expon exponen ents ts sum to ten, we note tha thatt   w2 x2 y2 z 2 =     10! w2 x2 y2 z 2 12 . Thus the coefficient must be 2, 2,10 2,2,2  = 2!2!2!2!2! 1.3.26.b. To make the exponents sum to 12, the coefficient must contain a factor of (−2)4 . Let  Let   a  = 2w, b  =  − According rding to the multino multinomial mial  −x, x, c  = 3y, d =  z  z,, e =  −2.  − 2. Acco   12   12! 2 2 2 2 4 theorem, the coefficient of  a  a b c d e = 2,2,2,2,4  = (2!) 4! , implying that the coefficient   12!   12! 2 2 2 4 of   w2 x2 y 2 z 2 = (2!) 4! 2 (−1) 3 (−2) = 2 4! 4 · 9 · 16. 4

4

4

1.3.26.c. 1.3.26 .c. In order for the exponen exponents ts to sum to 12, the coefficien coefficientt must contain contain a 4   12!   e12! 0 2 2 of 2 = 2 t 2 of  4 factor   v, = b =   w, c   =   −2x, d   =   y, e   = 5z, f  Th The coeffi coeffici cien ent a b c d2 e23f 4. isLe0t,  2a,212, 2=  wf   x2 y23. z 2 is (2!) 4! . Thus the coefficient of  w ,2,4 (2!) 4! (−2) 5 3 .





4

4

1.4.1.a. By applying    14!the formula for combinations with repetition, we see that the 10+5 1  = 10!4! . answer is 10 −

1.4.1.b. The num 1.4.1.b. number ber of wa ways ys to distr distribute ibute ten dimes among five childre children, n, when each child must receive at least one dime, is equal to the number 9 of  9!ways to distribute 5+5 1  = 5  = 4!5!  = 126. five dimes among five children, which is equal to 5 −

1.4.1.c. The number of ways to distribute ten dimes among five children, when the oldest must receive at least two dimes, is equal to the number of ways to distribute eight dimes among five children, which is equal to 8+5 1  = 12  =   12!  = 495. −

8

8

4!8!

1.4.7.a. By applying  35the formula for combinations with repetition, we see that the 32+4 1  = 32  = 6545. answer is 32 −

1.4.7.b. 32,,   where x where  x i  > 0  4  + x  + x  + x  >  0,, 1  ≤ i  ≤  i ≤  ≤ 4  x4  = 32  x3  +  x2  + x1  + ⇔   (x1 − 1) + (x (x2 − 1) + (x (x3 − 1) + (x (x4 − 1) = 28  =  xi − 1, 1  ≤ i  4 28,,   where y where  y i  = x  + y  + y  + y  ≤  i ≤  ≤ 4  y4  = 28  y3  +  y2  + ⇔   y1  + Note that in the last equation, each summand is positive  or zero. Thu Thuss the num number ber 31 28+4 1  = 4495  = of solutions is −



28

 28 3

 

1.4.7.c. x1  + x  +  x2  + x  +  x3  +  + x  x4  = 32 32,,   where x where  x 1 , x2  ≥  ≥ 5  5,, x3 , x4  ≥ 7  ≥  7 ⇔   (x1 − 5) + (x (x2 − 5) + (x (x3 − 7) + (x (x4 − 7) = 8 ⇔   y1  + y  +  y2  + y  +  y3  +  + y  y4  = 8,   where y where  y 1  =  = x  x1 − 5, y2  = x  =  x2 − 5, y3  = x  =  x3 − 7, y4  = x  =  x4 − 7 11 Note that the1 last equation, n, each summand is nonnegativ nonnegative. e. Thu Thuss the num number ber of   = 165  = equatio solution is in8+4 8 8



−

  

1.4.7.d. x1  +  + x  x2  +  + x  x3  +  + x  x4  = 32 32,,   where x where  x i  ≥ 8  ≥  8,, 1  ≤ i  ≤  i ≤  ≤ 4  4 ⇔   (x1 − 8) + (x (x2 − 8) + (x (x3 − 8) + (x (x4 − 8) = 0  =  xi − 8, 1 ≤ i  4 where  y i  = x  + y  + y  + y  ≤  i ≤  ≤ 4  y4  = 0,   where y  y3  +  y2  + ⇔   y1  + Note that in equation, nonnegative. e. Thu Thuss the num number ber of   n, each summand is nonnegativ 0+4the1lastequatio 3  = 0  = 1 solution is 0 −

1.4.7.e. x1  +  + x  x2  +  + x  x3  +  + x  x4  = 32 32,,   where x where  x i  ≥ −2, 1  ≤ i  ≤  i ≤  ≤ 4  4 ( x4  + 2) = 40 ( x3  + 2) + (x (x2  + 2) + (x ⇔   (x1  + 2) + (x ⇔   y1  +  + y  y2  +  + y  y3  +  + y  y4  = 40 40,,   where  where   yi  = x  =  xi  + 2, 2 , 1 ≤ i  ≤  i ≤  ≤ 4  4

Note that in the last equation, nonnegative. e. Thu Thuss the num number ber of   equatio 43 n, each summand is nonnegativ 40+4 1  = 40  = 12341 solution is 40 −

1.4.7.f. x1  + x  +  x2  + x  + x3  +  + x  x4  = 32 32,,   where  where   x1 , x2 , x3  > 0  >  0,, 0  < x4  ≤ 25  ≤  25 ⇔   (x1 − 1) + (x (x2 − 1) + (x (x3 − 1) + (x (x4 − 1) = 28 ⇔   y1  + y  +  y2  + y  + y3  +  + y  y4  = 28 28,,   where y where  y 1  =  = x  x1 − 1, y2  = x  =  x2 − 1, y3  = x  =  x3 − 1, y4  = x  =  x4 − 1 ⇒   y1 , y2 , y3  ≥  ≥ 0  0,, 0 ≤  ≤ y  y4  ≤  ≤ 24  24 To find all solutions of the original equation, we find all the solutions to the last equation for  for   yi   ≥   0, 1  ≤  i   ≤  4, and then subtract the number of solutions for which y4  = 25 25,, 26 26,, 27 27,, 28.  1   = 44 4495. If  Without an upper limit on   y4   the number of solutions is 28+4 28 y4   is fixed, then it is convenient to rewrite ourequation as   y1  + y  +  y2 + y  +  y3   = 28 − y4 . 30 y 28 y +3 1  = 28 y . Letting y Thus the number of solutions when y when  y4  is fixed is 28 y  4  y34 take 5 Letting the values values 25, 26, 27, and 28 tells us that the final answer is 4495 4495− − 3 − 2 − 1 − 20 −

− 4

− 4

4

−

− 4 − 4

 

 p 1 1 0 0

2.1.8.a. q    ¬q 1 0 0 1 1 0 0 1

 p 1 1 1 1 0 0 0 0

2.1.8.h. q r p  →  → q  q 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 1

p ∨ ¬q    ¬( p ∨ ¬q )   ¬ p   ¬( p ∨ ¬q )  → ¬ p 1 0 0 1 1 0 0 1 0 1 1 1 1 0 1 1 q  → r  →  r 1 0 1 1 1 0 1 1

p →  → r  r   ( p →  p  → q   q ) ∨ (q  → r  →  r)) [( p [( p →  → q   q ) ∨ (q  → r  →  r)] )] →  → (  ( p  p →  → r  r)) 1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1

2.1.10.  p q r p  →  → q  q p  →  → r  r q  → r  →  r p  →  → (  (q  q  → r  →  r)) ( p →  p  → q   q )  →  → (  ( p  p →  → r  r))   [ p (q r)] [( p 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 0 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 Since all the entries in the column corresponding to [ p   p   →   (q   →   r)]  )]   →   [( p [( p   →   q )   → ( p →  p  → r  r)] )] are 1’s, the statement is a tautology. →

 →

→

2.2.18.a. distributive law; inverse law; identity. 2.2.18.b. absorp 2.2.18.b. absorption; tion; substi substitutio tution; n; comm commutati utative; ve; distr distributiv ibutivee law law;; inv inverse erse;; identity; DeMorgan’s law 2.2.19.b.  p ∨ q ∨ q  ∨ (¬ p ∧ ¬q ∧ q  ∧ r )   ( p ∨ q ) ∨ (¬ p ∧ ¬q ∧ q  ∧ r) ( p ∨ q ) ∨ [¬( p ∨ q ) ∧ r ] [( p [( p ∨ q ) ∨ ¬( p ∨ q ))]] ∧ [( [( p  p ∨ q ) ∨ r] T 0 ∧ [( p ∨ q ) ∨ r ] ( p ∨ q ) ∨ r    p ∨ q ∨ q  ∨ r  

Reasons

Associative Law DeMorgan’s Law Dis Distri tribut butiv ivee La Law w Inverse Law Identity Associative Law

2.3.8. 1. premise 5

→

q)

→

( p → r)]

 

2. conjunctive simplification, (1) 2. premise 4. modus ponens, (2), (3) 5. conjunctive simplification, (1) 6. conjunction (4), (5) 7. premise 8. substituting an implication for its contrapositive, (7) 9. DeMorgan’s law, (8) 10. modus ponens (6), (9) 11. premise 12. substituting an implication for its contrapositive, (11) 13. DeMorgan’s law, double negation, (12) 14. modus ponens (10), (13) 14. conjunctive simplification (14) 2.3.10.a. 1. ( p ∧ ¬q ) ∧ r   2.   p ∧ (¬q ∧ q  ∧ r) 3.   p ∧ (r ∧ ¬q ) 4. ( p ∧ r) ∧ ¬q    5.   p ∧ r   6.   ∴  ( p  ( p ∧ r) ∨ q    2.3.10.b. 1.   p   2.   p  →  → q   q    3.   q    4.   ¬¬ ¬¬q  q    5.   ¬q ∨ q  ∨ r   6.   ∴  r  

premise as assoc socia iati tiv ve la law w (1 (1)) co comm mmut utat ativ ivee la law w (2 (2)) associative law (3) conjunctive simplification (4) disjunctive amplification (5)

premise premise modus ponens, (1), (2) double negat negation, ion, (3) premise disjunctive syllogism (4), (5)

2.3.10.c. 1.   p  →  → q   q    2.   ¬q    3.   ¬ p   4.   ¬r   5.   ¬ p ∧ ¬r   6.   ∴  ¬(  ¬ ( p ∨ r) 2.3.10.d. 1.   p  →  → q   q    2.   r  → ¬q    3.   r   4.   ¬q    5.   ∴  ¬ p  ¬ p  

premise premise modus tolle tollens, ns, (1), (2) premise conjunction, (3), (4) DeM DeMorg organ’ an’ss law, law, (5)

premise premise premise modus ponens, (2), (3) modus tolle tollens, ns, (1), (4)

2.3.10.e. 6

 

1.   p   2.   ¬¬ p   3.   ¬q  →  → ¬ p   4.   ¬¬ ¬¬q  q    5.   q    6.   p  →  → (  (q  q  → r  →  r)) 7.   q  → r  →  r   8.   ∴  r   2.3.10.f. 1.   p ∧ q    2.   p   3.   p  →  → (  (rr ∧ q ) 4.   r ∧ q    5.   r   6.   r  →  → (  (ss ∨ t) 7.   s ∨ t   8.   ¬s   9.   ∴  t   1.  2.3.10.g. ¬s   2.   p ∨ s   3.   p   4.   p  →  (q   → ( q  → r  →  r)) 5.   q  → r  →  r   6.   t →  → q   q    7.   t →  → r  r   8.   ∴  ¬r  ¬ r  → ¬t  

premise double negat negation, ion, (1) premise modus tolle tollens, ns, (2), (3) double negat negation, ion, (4) pr preemi mise se modus ponens, (1), (6) modus ponens, (5), (7) premise conjunctive simplification, (1) pr preemi mise se modus ponens, (2), (3) conjunctive simplification, (4) premise modus ponens, (5), (6) premise disjunctive syllogism, (7), (8) premise premise disjunctive syllogism, (1), (2) pr preemi mise se modus ponens, (3), (4) premise hypothetical syllogism, (5), (6) substitution, (7). (An implication is equiv equivalent alent to its contrapositive contrapositive.) .)

2.3.10.h. 1.   ¬r   premise 2.   ¬ p ∨ r   premise 3.    p¬ p∨ q      premise disjunctive syllogism, (1), (2) 4. 5.   ∴  q    disjunctive syllogism, (3), (4) 2.4.2.a.i.  p(3)  p(3) = true q (3) (3) = false ¬r(3) = false ⇒  p(3)  p(3) ∨ [q (3) (3) ∨ ¬r(3)] = false 2.4.2.a.ii.  p(2)  p(2) = true q (2) (2) = true r (2) = true ⇒  p(2)  p(2) →  → [ [q  q (2) → (2)  → r  r(2)] (2)] = true 7

 

2.4.2.a.iii. [ p(2) (2)]  → r (2) = true  p(2) ∧ q (2)] →  r(2) 2.4.2.a.iv.  p(0)  p(0) = true ¬ q (−=1)true = true r (1) ⇒  p(0)  p(0) →  → [ [¬ ¬q (−1) 1) ↔  ↔ r  r(1)] (1)] = true 2.4.2.b. The in 2.4.2.b. integer teger   x   must be strictly positive, even, and less than or equal to three ⇒ three  ⇒ x  x =  = 2. 2.4.8.a. true 2.4.8.b. false, x false,  x =  = 1 is a counterexample. 2.4.18.a.   ¬∃ ¬∃x x[ p(  p(x) ∨ q (x)] ⇔ ∀x¬[ p(  p(x) ∨ q (x)] ⇔ ∀x[¬ p(  p(x) ∧ ¬q (x)]

2.4.18.b.   ¬∀ ¬∀x x[ p(  p(x) ∧ ¬q (x)] ⇔ ∃x¬[ p(  p(x) ∧ ¬q (x)] ⇔ ∃x[¬ p(  p(x) ∨ ¬¬ ¬¬q  q (x)] ⇔ ∃x[¬ p(  p(x) ∨ q (x)] 2.4.24. Note that ∀ that  ∀x x∃y[x +  + y  y =  = 17] ⇔ 17]  ⇔ ∀x∃y[y  = 17 − x]. a. true. If  x  x  is an integer, then 17 − x  is also an integer. b. false.   x = 18 is a counterexample. c. false, as above. d. true. If   x  is a positive integer, then 17 − x  is an integer. change 4.1.2.a. Note that the assert 4.1.2.a. assertion ion ni=1 2i 1 = in=01  2i follows from  an simple 1 i = 2n − 1. of variabl ariables. es. We will use induc induction tion  to prov provee the last equal equality ity,, that i=0   2   that P (1) (1) asserts i0=0 2i = Let P  Let  P ((n) denote the statement in=01  2i = 2n − 1. Note that P  21 −   1, whi which ch is a tru truee sta state temen ment. t. We now now set out to pro prove ve that that   ∀k   ∈   N[P ( P (k )   ⇒ P  P ((k + 1)]. Assume P  Assume  P ((k ) is true. This implies that

 

−

 

−

−

−

k  i=0

k 1  −

i

2 =

2i + 2k = 2k − 1 + 2 k = 2 · 2k − 1 = 2k+1 − 1  ⇒  ⇒ P   P ((k  + 1) is true.

i=0

Note te ttha hatt  i(2i ) = 2 + (n − 1)2n+1 . No 4.1.2.b. 4.1.2. b. Let Let   P ( P   (n1) denote the assertion ni=1 i(2 1 2 P  P (1) (1) asserts that i=1 1(2 ) = 2 + (0)2 , which is a true statement. We now set out

 

8

 

to prove that ∀ that  ∀k k  ∈   N, [P ( P (k )  ⇒  ⇒ P   P ((k + 1)]. Assume P  Assume  P ((k ) is true. This implies that k+1 

i(2i ) =

k 

i(2i ) + (k (k + 1) · 2k+1

i=1

i=1

= = = = ⇒

2 + (k − 1)2k+1 + (k (k + 1) · 2k+1 2 + [(k − 1) + (k (k + 1)] · 2k+1 2 + 2k · 2k+1 2+ + k  k · 2k+2   P ( P (k + 1) is true.

(n + 1)! − 1. Not Notee tha thatt   P (1) 4.1.2.c. Let 4.1.2.c. Let   i(i!) = (n P (1) P (n) denote the assertion ni=1 i(  1  P ( asserts that i=1 1(1!) = 2! − 1, which is a true statement. We now set out to prove that ∀ that  ∀k k  ∈   N, [P  P ((k )  ⇒  ⇒ P   P ((k + 1)]. Assume P  Assume  P ((k ) is true. This implies that

 

k+1 

i(i!) =

k 

i(i!) + (k (k  + 1)[(k 1)[(k + 1)!]

i=1

i=1

= = = = ⇒

(k  + 1)! − 1 + (k ( k + 1)[(k 1)[(k + 1)!] (1 + k +  k  + 1)[(k 1)[(k + 1)!] − 1 (k  + 2)[(k 2)[(k + 1)!] − 1 (k  + 2)! − 1   P ( P (k + 1) is true.

2

 (n+1)

 i  = n  n 3 + (n (n + 1)3 . The equations 4.1.18. Let  Let   P ( P (n) denote the statement i=n +1 i = found in the book already verify that   P (0) P (0),, P (1) P (1),, P (2), P (2), and  and   P (3) P (3) are true. true. We can 2

b   i =a xi   has has b  b − a + 1 sum summa mand nds. s. So prove ∀ prove  ∀n n  ∈  N P  P ((n) directly. In general, the sum

our sum has (n (n + 1)2 − (n2 + 1) + 1 = n =  n 2 + 2n 2n + 1 − n2 − 1 + 1 = 2n + 1 summan summands. ds. Observe that (n+1)2



i  = (n2 +1) +(n +(n2 +2) +(n +(n2 +3)+ . . . + (n2 + 2n − 1)+( 1)+(n n2 + 2n) + (n2 + 2n + 1). 1).

i=n2 +1

Thus, each of the 2n 2n + 1 summands contains n contains  n 2 . Collecting these terms yields (n+1)2



i   = (2 (2n n + 1)n 1)n2 + 1 + 2 + 3 + . +  . . . + (2n (2n + 1)

i=n2 +1

2n+1 2

= (2 (2n n + 1)n 1)n +

 i=1

9

i

 

 (2  (2n n + 1)(2n 1)(2n + 1 + 1) 2  2(2n  2(2 n + 1)(n 1)(n + 1) = (2 (2n n + 1)n 1)n2 + 2 2 = (2 (2n n + 1)n 1)n + (2n (2n + 1)(n 1)(n + 1) = 2n3 + n2 + 2n 2n2 + 2n 2n + n  + n +  + 1

= (2 (2n n + 1)n 1)n2 +

3

3

2

=   n + (n (n + 3n 3n + 3n 3n + 1) =   n3 + (n (n + 1)3 (n+1)2 ∴



i   =   n3 + (n (n + 1)3

i=n2 +1

4.1.24.a.   a3  = 3, a4  = 5, a5  = 8, a6  = 13 13,, a7  = 21 4.1.24.b. Let 4.1.24.b. Let   an  defined as in the book and let   P ( P (n) denote the statement  statement   an   < n (7 (7//4) . The sta statem temen entt   P (1) P (1) asserts that 1   <   7/4, which which is tru true. e. We prov provee   ∀n   ∈ + Z P  P ((n) by induction. Assume P  Assume  P ((n) is true for all positive integers less than or equal to some fixed positive integer k integer  k.. ak+1  = a  =  a k +ak 1