Math 221 week 2 individule
Exercise 4.1
2. Establish each of the following for all n ≥ 1 by the Principle of Mathematical Induction. Solution
a)
S(n): ==,
S(1): = = =1, So S(1) is true.
Assume S(k): =
Consider S(k+1) = = +=-1+= -1. Hence, it follows that S(k)⇒S(k + 1) is true for all n ∈ Z+ by the Principle of Mathematical Induction.
b)
S( n) for n=1, = 2 = 2+(1-1). So S(1) is true.
Inductive Step: assume S(k)is true, for some (particular) k ∈ Z+—that is, assume that =2+(k-1). For n=k+1, = + (k+1) = 2+ (k-1)+(k+1)= 2+ (2k)= (2=k+1)= 2+k*. So by the Principle of Mathematical Induction, S(n) is true for all n ∈ Z+.
c)
For n=1, =1=(1+1)!-1. So S(1)is true. Assume S(k)is true, for some (particular) k ∈ Z+—that is, assume that = (k+1)!-1. For n=k+1, = +(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!=[ 1+(k+1)](k+1)!-1=(K+2)(k+1!-1=(k2)!-1. Hence, it follows that S(k)⇒S(k + 1) is true for all n ∈ Z+ and since S(1) is true for all problems a-c, n ≥ 1is true by the Principle of Mathematical Induction.
18. Consider the following four equations:
1) 1 = 1
2) 2 + 3 + 4 = 1 + 8
3) 5 + 6 + 7 + 8 + 9 = 8 + 27
4) 10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64
Conjecture the general formula suggested by these four equations, and proves your conjecture.
a) Conjecture: the general formula suggested by these four equations shows the sum of sequential integers from (n^2+1) to (n+1)^2 = n^3 + (n+1)^3. Therefore the Sum can written as for all of n∈N, (), (), ()+… +(n = =+ +(n
a. Proof: ==(2n+1)+(2n+1)(2n+2)/2=+(
4.2
5. Give a recursive definition for the intersection of the sets A1, A2, . . . , An, An+1 ⊆ _, n ≥ 1. Use the result in part (a) to show that for all n, r ∈ Z+ with n ≥ 3 and 1 ≤ r < n,(A1 ∩ A2 ∩ ・ ・ ・ ∩ Ar ) ∩ (Ar+1 ∩ ・ ・ ・ ∩ An) _ A1 ∩ A2 ∩ ・ ・ ・ ∩ Ar ∩ Ar+1 ∩ ・ ・ ・ ∩ An.
Solution
a) The intersection of , is ∩ .
The intersection of , , … ,
2. Establish each of the following for all n ≥ 1 by the Principle of Mathematical Induction. Solution
a)
S(n): ==,
S(1): = = =1, So S(1) is true.
Assume S(k): =
Consider S(k+1) = = +=-1+= -1. Hence, it follows that S(k)⇒S(k + 1) is true for all n ∈ Z+ by the Principle of Mathematical Induction.
b)
S( n) for n=1, = 2 = 2+(1-1). So S(1) is true.
Inductive Step: assume S(k)is true, for some (particular) k ∈ Z+—that is, assume that =2+(k-1). For n=k+1, = + (k+1) = 2+ (k-1)+(k+1)= 2+ (2k)= (2=k+1)= 2+k*. So by the Principle of Mathematical Induction, S(n) is true for all n ∈ Z+.
c)
For n=1, =1=(1+1)!-1. So S(1)is true. Assume S(k)is true, for some (particular) k ∈ Z+—that is, assume that = (k+1)!-1. For n=k+1, = +(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!=[ 1+(k+1)](k+1)!-1=(K+2)(k+1!-1=(k2)!-1. Hence, it follows that S(k)⇒S(k + 1) is true for all n ∈ Z+ and since S(1) is true for all problems a-c, n ≥ 1is true by the Principle of Mathematical Induction.
18. Consider the following four equations:
1) 1 = 1
2) 2 + 3 + 4 = 1 + 8
3) 5 + 6 + 7 + 8 + 9 = 8 + 27
4) 10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64
Conjecture the general formula suggested by these four equations, and proves your conjecture.
a) Conjecture: the general formula suggested by these four equations shows the sum of sequential integers from (n^2+1) to (n+1)^2 = n^3 + (n+1)^3. Therefore the Sum can written as for all of n∈N, (), (), ()+… +(n = =+ +(n
a. Proof: ==(2n+1)+(2n+1)(2n+2)/2=+(
4.2
5. Give a recursive definition for the intersection of the sets A1, A2, . . . , An, An+1 ⊆ _, n ≥ 1. Use the result in part (a) to show that for all n, r ∈ Z+ with n ≥ 3 and 1 ≤ r < n,(A1 ∩ A2 ∩ ・ ・ ・ ∩ Ar ) ∩ (Ar+1 ∩ ・ ・ ・ ∩ An) _ A1 ∩ A2 ∩ ・ ・ ・ ∩ Ar ∩ Ar+1 ∩ ・ ・ ・ ∩ An.
Solution
a) The intersection of , is ∩ .
The intersection of , , … ,