Question

SOMEONE PLEASE PLEASE PLEASE HELP!! Will fan and medal help!!

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) =




2. 12 + 42 + 72 + ... + (3n - 2)2 =




For the given statement Pn, write the statements P1, Pk, and Pk+1.
(2 points)
3. 2 + 4 + 6 + . . . + 2n = n(n+1)

Answer 1

i believe your copy/paste is missing information

Answer 2

Where?

Answer 3

1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) =
2. 12 + 42 + 72 + ... + (3n - 2)2 =

what does it equal? and the paste ate all of the ^s

Answer 4

That equal sign is to represent you starting to work out the equation

Answer 5

you cant prove the statements since they are incomplete

Answer 6

How are they incomplete?

Answer 7

prove that 3x =

Answer 8

prove that the color of the train is .........

Answer 9

prove that a tree is made of ........

Answer 10

there should be formulas attached to the right side of the equal sign in order to make a comparison with and to be able to reference to when formating the induction process

Answer 11

okay for number one it is equal to 4(4n+1)(8n+7)/6

Answer 12

well, lets prove that for n=1 :)

Answer 13

for number two it is equal to n(6n^2-3n-1)/2

Answer 14

Okay how would we do that?

Answer 15

replace n with a 1 ....

does: 4(1) (4(1)+2) = 4(4(1)+1)(8(1)+7)/6 ??

Answer 16

No.

Answer 17

then as long as you posted it correctly, then we have proven that this is not true for ALL positive integers since we know it is not true for at least one of them

Answer 18

Oh okay. So how would we solve number 2?

Answer 19

same first step; let n=1, is the setup true for n=1?

Answer 20

... + (3n - 2)2 = n(6n^2-3n-1)/2 ; let n=1

does: (3(1) - 2)2 = 1(6(1)^2 -3(1)-1)/2 ?

i believe thats spose to be a ^2 onthe left ....

Answer 21

Yes it is sorry

Answer 22

then we have a true condition: 1 = 1 for n=1

assume it is true for some n=k; and see if we can format it the same way for the induction step

Answer 23

How did you get 1=1 for n=1?

Answer 24

i just work the math ....

(3(1) - 2)^2 = 1(6(1)^2-3(1)-1)/2
(3 - 2)^2 = (6(1)-3-1)/2
(1)^2 = (6-4)/2
1 = 2/2 = 1


12 + 42 + 72 + ... + (3k - 2)^2 = k(6k^2-3k-1)/2

Answer 25

So how do we solve it now with k?

Answer 26

k is just a generic placeholder for "some value of n". If we can show that this same format is true for n=k+1, then we know that since its true for n=k=1, then its true for n=2, n=3, n=4, ...

Answer 27

putting it into the k format seems a little redundant, but its just for generality aspects

Answer 28

How would you write out the steps?

Answer 29

like i did, assume its true for some n=k

12 + 42 + 72 + ... + (3k - 2)^2 = k(6k^2-3k-1)/2

Answer 30

What would you do next?

Answer 31

let P(k) equal that left side for a clean up

P(k) = k(6k^2-3k-1)/2



now add the (k+1) element to each side

P(k) + (3(k+1) - 2)^2 = k(6k^2-3k-1)/2 + (3(k+1) - 2)^2


the left side is equal to P(k+1); can we form the right side of it into a k+1 version as well?
remember, all we know is a format; so we have to show that the format is consistent

Answer 32

\[ \frac12k(6k^2-3k-1) + (3(k+1) - 2)^2\]
\[ \frac12(6k^3-3k^2-k) + 9(k+1)^2 +4-12(k+1)\]
\[ \frac12(6k^3-3k^2-k+ 18(k+1)^2 +8-24(k+1)) \]
work the math .....

Answer 33

typoed it .... we want to try to get it into the format:
\[\frac12(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 34

Okay I get it now. How would you do number 3?

Answer 35

its the same process, let n=1, let n=k, then proof it out for n=k+1

Answer 36

So for n=1 it you would end up having 2(1)=1(1+1)?

Answer 37

correct

Answer 38

Okay. Thank you so much.

Answer 39

youre welcome