Question

PLEASE HELP AND SHOW ALL WORK

7.04

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
(4 points each.)

1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = quantity four times quantity four n plus one times quantity eight n plus seven divided all divided by six


2. 12 + 42 + 72 + ... + (3n - 2)2 = quantity n times quantity six n squared minus three n minus one all divided by two


For the given statement Pn, write the statements P1, Pk, and Pk+1.
(2 points)

2 + 4 + 6 + . . . + 2n = n(n+1)

Answer 1

4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = 4(4n+1)(8n+7)/6 is false because it isn't even true when n=1. 4 ⋅ 6 = 4(4*1+1)(8*1+7)/6 24 = 4(5)(15)/6 24 = 50 Also 4n(4n+2) is not even the correct formula for the nth term. The correct nth term formula of 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... is (n+3)(n+5). So it's wrong all the way around.

Answer 2

1. False.

2.False.

3.
`P_(1) =`2

`P_(k)`=2+4+6+.....+2n= n(n+1)

`P_(k+1)`=2+4+6+....+2(k+1)=(k+1)(k+2)

Explanation:

1.Given

4.6+5.7+6.8+.....+4n(4n+2)=
`(4(4n+1)(8n+7))/(6)`

First , we prove for n=1

`P_(1)`= 4.6=24

Left hand side :

`P_(1)`=
`4*1(4*1+2)`

`P_(1)`=24

Right hand side:

`P_(n)`=
`(4(4n+1)(8n+7))/(6)`

Put n=1

`P_(1)`=
`(4(4*1 +1)(8*1 +7))/(6)`

`P_(1)`=
`(4(4*1 +1)(8*1+7))/(6)`

`P_(1)`=
`(4*5*15)/(6)`

`P_(1)`=50

Hence, left hand side≠ right hand side

Therefore, the given statement is false.

2.
`P_(n)`=12+42+72+.......(3n-2)2=
`(n(6n^(2-3n-1)) )/(2)`

similarly , we prove that in the same manner

First , we prove for n=1 the given statement is true or false.

Take n=1

`P_(1)`=12

Left hand side :

`P_(n)`=(3n-2)2

Put n=1 then

`P_(1)`=
`(3*1-2)2`

`P_(1)`=2

Right hand side:

`P_(n) =[tex](n(6n^(2-3n-1)) )/(2)`

Put n=1

Then we get

`P_(1)`=
`(1(6*1-3*1-1))/(2)`

`P_(1)`=
`(4)/(2)`

`P_(1)`=2

Hence , left hand side≠right hand side ≠
`P_1`

Therefore, the given statement is false.

3. Given that

`P_(n)`=2+4+6+....+2n=n(n+1)

Put n= 1

`P_(1)`=2

Put n=k then we get the statement Pk

`P_(k)`=2+4+6+....+2k=k(k+1)

Now , Put n= k+1 and then we get the statment Pk+1

`P_(k+1)`=2+4+6+....+2(k+1)=(k+1)(k+2).