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Solutions to Exercises on Mathematical Induction Math 1210 ...

Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cIn Exercises 1-15 use Mathematical Induction to establish the formula forn + 22+ 32+ +n2=n(n+ 1)(2n+ 1)6 Proof:Forn= 1, the statement reduces to 12=1 2 36and is obviously the statement is true forn=k:12+ 22+ 32+ +k2=k(k+ 1)(2k+ 1)6,(1)we will prove that the statement must be true forn=k+ 1:12+ 22+ 32+ + (k+ 1)2=(k+ 1)(k+ 2)(2k+ 3)6.(2)The left-hand side of (2) can be written as12+ 22+ 32+ +k2+ (k+ 1) view of (1), this simplifies to:(12+ 22+ 32+ +k2)+ (k+ 1)2=k(k+ 1)(2k+ 1)6+ (k+ 1)2=k(k+ 1)(2k+ 1) + 6(k+ 1)26=(k+ 1)[k(2k+ 1) + 6(k+ 1)]6=(k+ 1)(2k2+ 7k+ 6)6=(k+ 1)(k+ 2)(2k+ 3) the left-hand side of (2) is equal to the right-hand side of (2).

Thus the left-hand side of (2) is equal to the right-hand side of (2). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integer n. 2. 3 + 32 + 33 + + 3n = 3n+1 3 2 Proof: For n = 1, the statement reduces to 3 = 32 3 2

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Transcription of Solutions to Exercises on Mathematical Induction Math 1210 ...

1 Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cIn Exercises 1-15 use Mathematical Induction to establish the formula forn + 22+ 32+ +n2=n(n+ 1)(2n+ 1)6 Proof:Forn= 1, the statement reduces to 12=1 2 36and is obviously the statement is true forn=k:12+ 22+ 32+ +k2=k(k+ 1)(2k+ 1)6,(1)we will prove that the statement must be true forn=k+ 1:12+ 22+ 32+ + (k+ 1)2=(k+ 1)(k+ 2)(2k+ 3)6.(2)The left-hand side of (2) can be written as12+ 22+ 32+ +k2+ (k+ 1) view of (1), this simplifies to:(12+ 22+ 32+ +k2)+ (k+ 1)2=k(k+ 1)(2k+ 1)6+ (k+ 1)2=k(k+ 1)(2k+ 1) + 6(k+ 1)26=(k+ 1)[k(2k+ 1) + 6(k+ 1)]6=(k+ 1)(2k2+ 7k+ 6)6=(k+ 1)(k+ 2)(2k+ 3) the left-hand side of (2) is equal to the right-hand side of (2).

2 Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive + 32+ 33+ + 3n=3n+1 32 Proof:Forn= 1, the statement reduces to 3 =32 32and is obviously the statement is true forn=k:3 + 32+ 33+ + 3k=3k+1 32,(3)we will prove that the statement must be true forn=k+ 1:3 + 32+ 33+ + 3k+1=3k+2 32.(4) Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cThe left-hand side of (4) can be written as3 + 32+ 33+ + 3k+ 3k+ view of (3), this simplifies to:(3 + 32+ 33+ + 3k)+ 3k+1=3k+1 32+ 3k+1=3k+1 3 + 2 3k+12=3 3k+1 32=3k+2 the left-hand side of (4) is equal to the right-hand side of (4).

3 Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive + 23+ 33+ +n3=n2(n+ 1)24 Proof:Forn= 1, the statement reduces to 13=12 224and is obviously the statement is true forn=k:13+ 23+ 33+ +k3=k2(k+ 1)24,(5)we will prove that the statement must be true forn=k+ 1:13+ 23+ 33+ + (k+ 1)3=(k+ 1)2(k+ 2)24.(6)The left-hand side of (6) can be written as13+ 23+ 33+ +k3+ (k+ 1) view of (5), this simplifies to:(13+ 23+ 33+ +k3)+ (k+ 1)3=k2(k+ 1)24+ (k+ 1)3=k2(k+ 1)2+ 4(k+ 1)34=(k+ 1)2[k2+ 4(k+ 1)]4=(k+ 1)2(k+ 2) the left-hand side of (6) is equal to the right-hand side of (6).

4 Thisproves the inductive step. Therefore, by the principle of mathematicalSolutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cinduction, the given statement is true for every positive + 3 + 6 + 10 + +n(n+ 1)2=n(n+ 1)(n+ 2)6 Proof:Forn= 1, the statement reduces to 1 =1 2 36and is obviously the statement is true forn=k:1 + 3 + 6 + 10 + +k(k+ 1)2=k(k+ 1)(k+ 2)6,(7)we will prove that the statement must be true forn=k+ 1:1 + 3 + 6 + 10 + +(k+ 1)(k+ 2)2=(k+ 1)(k+ 2)(k+ 3)6.

5 (8)The left-hand side of (8) can be written as1 + 3 + 6 + 10 + +k(k+ 1)2+(k+ 1)(k+ 2) view of (7), this simplifies to:[1 + 3 + 6 + 10 + +k(k+ 1)2]+(k+ 1)(k+ 2)2=k(k+ 1)(k+ 2)6+(k+ 1)(k+ 2)2=k(k+ 1)(k+ 2) + 3(k+ 1)(k+ 2)6=(k+ 1)(k+ 2)(k+ 3) the left-hand side of (8) is equal to the right-hand side of (8). Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive + 4 + 7 + + (3n 2) =n(3n 1)2 Proof:Forn= 1, the statement reduces to 1 =1 22and is obviously the statement is true forn=k:1 + 4 + 7 + + (3k 2) =k(3k 1)2,(9)we will prove that the statement must be true forn=k+ 1:1 + 4 + 7 + + [3(k+ 1) 2] =(k+ 1)[3(k+ 1) 1]2.

6 (10) Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cThe left-hand side of (10) can be written as1 + 4 + 7 + + (3k 2) + [3(k+ 1) 2].In view of (9), this simplifies to:[1 + 4 + 7 + + (3k 2)] + (3k+ 1) =k(3k 1)2+ (3k+ 1)=k(3k 1) + 2(3k+ 1)2=3k2+ 5k+ 22=(k+ 1)(3k+ 2) last expression is obviously equal to the right-hand side of (10). Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive + 32+ 52+ + (2n 1)2=n(2n 1)(2n+ 1)3 Proof:Forn= 1, the statement reduces to 12=1 3 33and is obviously the statement is true forn=k:12+ 32+ 52+ + (2k 1)2=k(2k 1)(2k+ 1)3,(11)we will prove that the statement must be true forn=k+ 1:12+ 32+ 52+ + [2(k+ 1) 1]2=(k+ 1)[2(k+ 1) 1][2(k+ 1) + 1]3.

7 (12)The left-hand side of (12) can be written as12+ 32+ 52+ + (2k 1)2+ [2(k+ 1) 1] view of (11), this simplifies to: Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi c[12+ 32+ 52+ + (2k 1)2]+ (2k+ 1)2=k(2k 1)(2k+ 1)3+ (2k+ 1)2=k(2k 1)(2k+ 1) + 3(2k+ 1)23=(2k+ 1)[k(2k 1) + 3(2k+ 1)]3=(2k+ 1)(2k2+ 5k+ 3)3=(2k+ 1)(k+ 1)(2k+ 3)3=(k+ 1)[2(k+ 1) 1][2(k+ 1) + 1] the left-hand side of (12) is equal to the right-hand side of (12). Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive + 5 + 9 + 13 + + (4n 3) = 2n2 nProof:Forn= 1, the statement reduces to 1 = 2 12 1 and is obviously the statement is true forn=k:1 + 5 + 9 + 13 + + (4k 3) = 2k2 k ,(13)we will prove that the statement must be true forn=k+ 1:1 + 5 + 9 + 13 + + [4(k+ 1) 3] = 2(k+ 1)2 (k+ 1).

8 (14)The left-hand side of (14) can be written as1 + 5 + 9 + 13 + + (4k 3) + [4(k+ 1) 3].In view of (13), this simplifies to:[1 + 5 + 9 + 13 + (4k 3)] + (4k+ 1) = (2k2 k) + (4k+ 1)= 2k2+ 3k+ 1 = (k+ 1)(2k+ 1)= (k+ 1)[2(k+ 1) 1]= 2(k+ 1)2 (k+ 1).Thus the left-hand side of (14) is equal to the right-hand side of (14). Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive to Exercises on Mathematical InductionMath 1210, Instructor: M.

9 Despi + 23+ 25+ + 22n 1=2(22n 1)3 Proof:Forn= 1, the statement reduces to 2 =2(22 1)3and is obviously the statement is true forn=k:2 + 23+ 25+ + 22k 1=2(22k 1)3,(15)we will prove that the statement must be true forn=k+ 1:2 + 23+ 25+ + 22(k+1) 1=2(22(k+1) 1)3.(16)The left-hand side of (16) can be written as2 + 23+ 25+ + 22k 1+ 22(k+1) view of (15), this simplifies to:(2 + 23+ 25+ + 22k 1)+ 22k+1=2(22k 1)3+ 22k+1=2(22k 1) + 3 22k+13=22k+1 2 + 3 22k+13=4 22k+1 23=2(2 22k+1 1)3=2(22k+2 1) last expression is obviously equal to the right-hand side of (16).

10 Thisproves the inductive step. Therefore, by the principle of mathematicalinduction, the given statement is true for every positive 2 3+12 3 4+13 4 5+ +1n(n+ 1)(n+ 2)=n(n+ 3)4(n+ 1)(n+ 2)Proof:Forn= 1, the statement reduces to11 2 3=1 44 2 3and is the statement is true forn=k:11 2 3+12 3 4+13 4 5+ +1k(k+ 1)(k+ 2)=k(k+ 3)4(k+ 1)(k+ 2),(17) Solutions to Exercises on Mathematical InductionMath 1210, Instructor: M. Despi cwe will prove that the statement must be true forn=k+ 1:11 2 3+12 3 4+13 4 5+ +1(k+ 1)(k+ 2)(k+ 3)=(k+ 1)(k+ 4)4(k+ 2)(k+ 3).