Question

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

4 * 6 + 5 * 7 + 6 * 8 + ... + 4n(4n + 2) =
4(4n + 1)(8n + 7) {over} 6

Answer 1

try this for n=1

Answer 2

do I put it in the fraction?

Answer 3

wait, okay I got 24 when I subbed it in and got 24, the same number as when you multiply the first two numbers @TuringTest

Answer 4

4n(4n + 2) = 4(4n + 1)(8n + 7)/6 ; try n=1, assuming your book says n starts at 1

4(6) = 4(5)(15)/6

4(6)(6) = 4(5)(15)

4(2.3)(2.3) = 4(5)(5.3)

(2)(3)(2) = (5)(5) doesnt seem to work for n=1 does it?

Answer 5

no it doesn't, oh okay i was supposed to put the one in the two parts equal to each other

Answer 6

well, for n=1 we only have those parts yes

and it turns out that in order for it to actually work, in the basis step, n has to be a complex number :)

Answer 7

so this statement is false because they don't equal each other?

Answer 8

correct, there is at least one value of n that this simply does not hold true. And since it is not true for ALL cases, it either needs to be amended or tossed out.

Answer 9

Okay I have a question,

4(6) = 4(5)(15)/6

4(6)(6) = 4(5)(15) <<<---What did you do to get from here

4(2.3)(2.3) = 4(5)(5.3) <<<---To here? Divide by six?

Answer 10

multiplied by 6; but that is immaterial. I was simply reworking it to its common factors to show that they do not exist.

Answer 11

simply letting n =1 to get 24 = 50 is an absurdity enough to claim its false

Answer 12

okay got it.

so for this,

1^2 + 4^2 + 7^2 + ...+ (3n - 2)^2 = n(6n^2 - 3n - 1)

I would take the two parts equal to each other and put one in for n?

Answer 13

n=1 is the basis step, if its is true for n=1, we can then show that it is true for n=2, 3, 4, 5, ... inductively

so, establish that (3n - 2)^2 = n(6n^2 - 3n - 1) is true for n=1

Answer 14

true or false that is; if true, there is more work to do :)

Answer 15

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

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

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

1^2 = 1(36 - 3 - 1)/2

= 1(32)/2

= 32/2

1^2 = 16


Seems like it doesn't work, unless I did something wrong...

Answer 16

theres a /2?

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

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

1 = (3 - 1 )/2

1 = 2/2 , then its true for n=1

Answer 17

yeah 6(1)^2 is not equal to (6)^2 :)

Answer 18

now to make it more general, let n=k and rewrite the setup, this way we can avoid n=1 for the moment and assume it is true for some n=k.


1^2 + 4^2 + 7^2 + ...+ (3k - 2)^2 = k(6k^2 - 3k - 1)/2

add on the next (k+1) formula to each side ...

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

now algebra the right hand side to LOOK like the original setup, but with (k+1):


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

if we can work it back to look like this, then we know that it works for some k, and for some k+1 .... always. we know it works for k=n=1, therefore it has to work for n=2,3,4,5,...

Answer 19

so .. get to work :)

k(6k^2 - 3k - 1)/2 + (3(k+1) - 2)^2, transformed into

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

Answer 20

I'm sorry, I'm just confused, I got that n=k, then from there,

I get...(3k - 2)^2 = k(6k^2 - 3k - 1)/2

Then what do I do with this (k + 1)

Answer 21

@amistre64

Answer 22

ill be right back ... got to change computers

Answer 23

ok

Answer 24

k+1 is what we need to prove; so we need to add the next "kth" value to both sides of the setup. what you add to one side you add to the other.

we know that for each new term on the left, it follows the expression:

(3k - 2)^2

so, the (k+1)th term has the same form, but with k+1 instead of k.

Answer 25

If you could do one line at a time that would be good

Answer 26

so, to prove it is good for the (k+1)th term; add

(3(k+1) - 2)^2

to each side

Answer 27

gotta move to the library, getting kicked out of my office space :)

Answer 28

lol okay

Answer 29

so then I get, (3(k + 1) - 2)^2 + (3k - 2)^2 = k(6k^2 - 3k - 1)/ + (3(k + 1) -2)^2

Answer 30

you almost see it .....

we already have it setup that:

1^2 + 4^2 + 7^2 + ...+ (3k - 2)^2 = k(6k^2 - 3k - 1)/2


to clear the clutter, lets shove that left hand side into some P(k)

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


now, P(k+1) = P(k) + (3(k+1) - 2)^2


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

Answer 31

all we really can say about this is that the "FORM" is correct:

[*] (6[*]^2 - 3[*] - 1)
-------------------
2

we want to show the P(k+1) fits that form

Answer 32

\[k\frac{(6k^2 - 3k - 1)}{2} + (3(k+1) - 2)^2\]
\[\frac{k(6k^2 - 3k - 1) + 2(3(k+1) - 2)^2}{2}\]
\[\frac{6k^3 - 3k^2 - k + 2(3k+1)^2}{2}\]
\[\frac{6k^3 - 3k^2 - k + 2(9k^2+6k+1)}{2}\]
\[\frac{6k^3 - 3k^2 - k + 18k^2+12k+2}{2}\]
\[\frac{6k^3 + 15k^2+11k+2}{2}\]
we need to factor out a k+1 from the top now , if possible

6k^3 + 15k^2+11k+2
0 -6 -9 -2
----------------------
-1 | 6 9 2 0

\[\frac{(k+1)(6k^2 + 9k+2)}{2}\]
dbl chk my work, im sure i messed up someplace along the way

Answer 33

in the second line where did you get the 2?

Answer 34

adding fractions .... 1/2 + a = (1+2a)/2

Answer 35

it might be helpful to try to get the finished outcome to this stage :) that way we can see a clearer way to get to it

Answer 36

\[\frac{(k+1)(6(k+1)^2 - 3(k+1) - 1)}{2} \]
\[\frac{(k+1)(6(k^2+2k+1) - 3k-3 - 1)}{2} \]
\[\frac{(k+1)(6k^2+12k+6 - 3k-4)}{2} \]
\[\frac{(k+1)(6k^2+9k+2)}{2} \]

Answer 37

thats was simpler lol soo, now we have a path to get from that stumper to the form we want ...

Answer 38

\[\frac{(k+1)(6k^2+9k+2)}{2}\]
\[\frac{(k+1)(6k^2+(12-3)k+(6-4))}{2}\]
\[\frac{(k+1)(6k^2+12k -3k+6-4)}{2}\]
\[\frac{(k+1)(6(k^2+2k +1)-3k-3-1))}{2}\]
\[\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
tada!! lol

Answer 39

all this does, is show us that if it works for some integer, k; then it also works for the k+1 integer as well.

we know it works for k=n=1; so it has to work for 1+1, for 1+1+1, for 1+1+1+1 .... etc

Answer 40

third line might be better if we swap the -3k and +6 to show that we then complete the square .... had it in my head :)

Answer 41

i have to go sit in class for 3 hours, so have fun with it ... think over how i did it and work it out in your own mind. good luck

Answer 42

thanks