fix

Question# 3 from Chapter 5.1 exercises

Let  P(n) be the statement that 12 + 22 + ...+n2 = n (n + 1)(2n + 1)/6 for the positive integer n.

A) What is the statement P(1) ?

P(1) = 12 =1

 

B) Show that P(1) is true, completing the basis step of the proof.

P(1) of n (n + 1)(2n + 1)/6 = 1 hence P(1) is true since it is equal to p(1) on the left

 

C) What is the inductive hypothesis?

We assume that P(k) holds for an arbitrary positive integer K i.e  12 + 22 + ...+k2 = k (k + 1)(2k + 1)/6

 

D) What do you need to prove in the inductive step?

We have to prove that P(k +1) is also true, i.e. 12 + 22 + ...+k2 + (k+1)2 = (k+1)(k+2)(2k+3)/6

=2k3+9k2+13k+6/6 also holds

 

E) Complete the inductive step, identifying where you use the inductive hypothesis

From 12 + 22 + ...+k2 = k (k + 1) (2k + 1)/6, if we add (k+1)2 on both sides we yield

 12 + 22 + ...+k2 + (k+1)2 = k (k+1) (k+2)/6 + (k+1)2

= (2k3 +3k2 +k) + 6 (k2 + 2k + 1)/6

= 2k3 + 9k2 +13K + 6/6

 

F) Explain why these steps show that this formula is true whenever n is a positive integer.

By mathematical induction we know that p (n) is true for all non-negative integers

 

This is question # 3

See below what you wrote in black that I highlighted and what I changed in blue.

E) Complete the inductive step, identifying where you use the inductive hypothesis

From 12 + 22 + ...+k2 = k (k + 1) (2k + 1)/6, if we add (k+1)2 on both sides we yield

 12 + 22 + ...+k2 + (k+1)2 = k (k+1) (k+2)/6 + (k+1)2  k(k+1)(2k+1)/6 + (k+1)2

= (2k3 +3k2 +k) + 6 (k2 + 2k + 1)/6    [k (k+1) (2k+1) + 6(k+1)2]/6

= 2k3 + 9k2 +13K + 6/    (k+1)[k(2k+1) + 6(k+1)]/6

=  (k+1)[2k2+k + 6k+6]/6

=  (k+1)(2k2 + 7k + 6)/6

=  (k+1)(2k+ 3)(k + 2)/6 = (k+1)[2(k+ 1) + 1][(k + 1) + 1]/6 

Let n = k + 1 to get the following and complete the proof.

RHS = n(2n + 1)(n + 1)/6 = n(n + 1)(2n + 1)/6

Question# 4 from Chapter 5.1 exercises **I was told to try this one again it was also wrong*

 Let, P(n) be the statement that 13 + 23 + ...+n3 = (n (n + 1)/2)2 for the positive integer n.

A) What is the statement P(1) ?

P(1) = 13 =1

 

B) Show that P(1) is true, completing the basis step of the proof.

P(1) of (n(n+1)/2)2 = (1(1+1)/2)2 = 12 = 1 hence P(1) is true since it is equal to P(1)  of the left hand side

 

C) What is the inductive hypothesis?

We assume that P(k) holds for an arbitrary positive integer k.  13 + 23 + ...+k3 = (k(k+1)/2)2

 

D) What do you need to prove in the inductive step?

We have to prove that p(k+1) is true.  i.e.  .  13 + 23 + ...+k3 + (k+1)3 =((k+1)(k+2)/2)2 is true

=k4+6k3+13k2+12k+4/4

 

E) Complete the inductive step, identifying where you use the inductive hypothesis

Adding (k+1)3 to both sides of P(k) we got   13 + 23 + ...+k3 + (k+1)3 =(k(k+1)/2)2 + (k+1)3

=k4 +2k3+k2/4 +k3 +3k2 +3k + 1

=k4 + 6k3 + 13k2 + 12k + 4/4 hence proved

 

F) Explain why these steps show that this formula is true whenever n is a positive integer.

By mathematical induction we know that p (n) is true for all non-negative integers

 

https://phoenix.vitalsource.com/#/books/9781308260235/cfi/432!/4/[email protected]:0.00