Let P(n) be the statement that 12 + 22 + ... + n2 = n 6(n + 1)(2n + 1) for positive integer n. (a) What is the statement P(1) ? (b) Show that P(1) is true, completing the basis step of the proof. (c) What is the inductive hypothesis? (d) What do you need to prove in the inductive step? (e) Complete the inductive step, identifying where you use the inductive hypothesis. (f) Explain why these steps show that this formula is true whenever n is a positive integer?

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.

a)P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$

=(1*2*3)/6=1

b) P(1)= $1^{2}=(1*(1+1)*(2(1)+1)))))/6$ ==(1*2*3)/6=1

P(1)=12

=1

c)Induction hypothesis : Assume for all positive integers n,

P(n)=$1^{2}+2^{2}+....n^{2}=(n(n+1)(2n+1))/6$ is true.

d)P(n+1)=$1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n+1)(n+2)(2n+3))/6$

e)Need to prove P(n+1) is true for Inductive step

Consider $1^{2}+2^{2}+....n^{2}+(n+1)^{2}=(n(n+1)(2n+1))/6 +(n+1)^{2}$

Focusing on right side ,

=(n(n+1)(2n+1)+6(n+1)2 )/6

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

=(n+1)(2n2+7n+6)/6

Use +4,+3 as factors for 2n2+7n+6 to get

=

((n+1)(n+2)(2n+3))/6 which is equal to P(n+1).

Hence P(n+1) is true.

Completing inductive step: Add n+1 term to the right side of P(n) to

show that it is equal to the RHS of P(n+1)

f. as whenever the statement holds for P(n) it hold for P(n+1) we have completed both the basis step and the inductive step , so by the principle of mathematical induction the statement is true for every positive integer n.