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## Here's the Solution to this Question

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))/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))/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))/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))/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))/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))/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))/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.