Question

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

Answer 1

2. \[12 + 42 + 72 + ... + (3n - 2)2 =\frac{ n(6n ^{2}-3n-1) }{ 2 }\]

Answer 2

We must first prove that this is true for n=1
(3(1)-2)2=1(6(1)^2-3(1)-1)/2

Answer 3

1=1 so it is true for n=1, right?

Answer 4

Yes correct, now we must assume that it is true for n=k, which is a random positive integer.
We cant prove this one correct, we are just assuming
(3k-2)^2=k(6k^2-3k-1)/2

Answer 5

And the last step is to then prove it is true for k+1

Answer 6

Okay, how do I do that? Sorry. I really bad at mathematical induction!

Answer 7

I'm*

Answer 8

Actually for the last step im going to simplify it more since we can first. k(6k^2-3k-1)/2=
(6k^3-3k^2-k)/2
so we are assuming that (3k-2)^2=(6k^3-3k^2-k)/2
now to do this, we are going to add another term to the left side
(3k-2)^2+(3(k+1)-2)^2=(k+1)(6(k+1)^2-3(k+1)-1/2
(3k-2)^2+(3k+3-2)^2=(k+1)(6(k+1)^2-3(k+1)-1/2
now before we go any further, we dont want to simplify the (3k-2)^2 because we are already assuming that is = (6k^3-3k^2-k)/2 so we can substitute this after simplifying everything else.
(3k-2)^2+(3k+1)^2=(k+1)(6(k^2+2k+1)-3k-3-1)/2
(3k-2)^2+9k^2+6k+1=(k+1)(6k^2+12k+6-3k-4)/2
(3k-2)^2+9k^2+6k+1=(k+1)(6k^2+9k+2)/2
(3k-2)^2+9k^2+6k+1=(6k^3+9k^2+2k+6k^2+9k+2)/2
(3k-2)^2+9k^2+6k+1=(6k^3+15k^2+11k+2)/2
(3k-2)^2+9k^2+6k+1=3k^3+7.5k^2+5.5k+1
now we out of simplifying, we substitute (6k^3-3k^2-k)/2 to the other side
(6k^3-3k^2-k)/2=3k^3-1.5k^2-.5k
3k^3-1.5k^2-.5k+9k^2+6k+1=3k^3+7.5k^2+5.5k+1
3k^3+7.5k^2+5.5k+1=3k^3+7.5k^2+5.5k+1
this is exactly what we were supposed to get!! its proven,
sorry that took so long.. you could go through my work to see if it makes sense but most of it is just simplifying, the substitution step is a little confusing, thats what I had trouble with when i learned this, and adding the extra term on the left

Answer 9

o.o

Answer 10

okay and so now its proven and i don't have to do anything else?

Answer 11

Yes its proven, by mathematical induction is true for n=1, k which is a random number and k+1, which means every number after that is also true

Answer 12

okay, thank you! :)

Answer 13

Actually for the last step im going to simplify it more since we can first.\(k(6k^2-3k-1)/2=
(6k^3-3k^2-k)/2\)

so we are assuming that \((3k-2)^2=(6k^3-3k^2-k)/2\)
now to do this, we are going to add another term to the left side

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

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

now before we go any further, we dont want to simplify the \((3k-2)^2\)

because we are already assuming that is = \((6k^3-3k^2-k)/2\) so we can substitute this after simplifying everything else.

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

\((3k-2)^2+9k^2+6k+1=(k+1)(6k^2+12k+6-3k-4)/2\)

\((3k-2)^2+9k^2+6k+1=(k+1)(6k^2+9k+2)/2\)

\((3k-2)^2+9k^2+6k+1=(6k^3+9k^2+2k+6k^2+9k+2)/2\)

\((3k-2)^2+9k^2+6k+1=(6k^3+15k^2+11k+2)/2\)

\((3k-2)^2+9k^2+6k+1=3k^3+7.5k^2+5.5k+1\)

now we out of simplifying, we substitute \((6k^3-3k^2-k)/2\) to the other side

\((6k^3-3k^2-k)/2=3k^3-1.5k^2-.5k\)

\(3k^3-1.5k^2-.5k+9k^2+6k+1=3k^3+7.5k^2+5.5k+1\)

\(3k^3+7.5k^2+5.5k+1=3k^3+7.5k^2+5.5k+1\)

this is exactly what we were supposed to get!! its proven,
sorry that took so long.. you could go through my work to see if it makes sense but most of it is just simplifying, the substitution step is a little confusing, thats what I had trouble with when i learned this, and adding the extra term on the left


@Venny use latex better ❤_❤

Answer 14

Please, I don't know latex :P

Answer 15

write like this
`\( your math stuff \)`

Answer 16

thank you, i will try ❤

Answer 17

+ awesome job though
u r smart
veeery smart one ❤_❤