Question

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

1. 4•6+5•7+6•8+...+4n(4n+2)=(4(4n+1)(8n+7)) / (6)

Answer 1

so you know that mathematical induction has four important steps to follow, right?

Answer 2

no please explain

Answer 3

@Percy* @BAdhi @thomas5267

Answer 4

How is this even right?
Take n=1
LHS=4x6=24
RHS=4x5x15/6=300/6=50

Answer 5

I think the statement is incorrect.

Answer 6

yeah is incorrect but you need to mention that, but the principles of mathematical induction, 1. 4•6+5•7+6•8+...+4n(4n+2)=(4(4n+1)(8n+7)) / (6) is not true for all integer n

Answer 7

it just has ? marks everywhere on your response @Percy*

Answer 8

and can y'all help me with one more?

Answer 9

You don't even need the principle of mathematical induction. If LHS does not equal RHS for one n, then the whole thing is invalid.

Answer 10

sorry, i dont know where this ? is coming from but 4get ?, coz look the equation... thomas i think we do becasue the question ask us to

Answer 11

Here is my second one, you do the same thing.

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

Answer 12

yeah, but what is the restriction say

Answer 13

what?

Answer 14

remember that our restriction for the first one says for all positive integers, that gave us a clue where to start

Answer 15

the second one has the exact same instructions.

Answer 16

yeah, you do the same thing

Answer 17

will you type it out please

Answer 18

or show me what i'm supposed to write

Answer 19

oky, this will b long, coz we will do step by step

Answer 20

ok but last time your response had a lot of ?'s in it because I guess you used symbols that my computer wasnt familiar with or something

Answer 21

no dont worry

Answer 22

so we have a basis step, inductive hypothesis, inductive step and conclusions

Answer 23

ok

Answer 24

guys i just typed somthing and doesnt show up....

Answer 25

ugh

Answer 26

let me try again

Answer 27

basis step( let n=1)\[1^2+4^2+7^2+...(3n-2)=\frac{ n(6n^2-3n-1) }{ 2}\]
\[lhs=(3n-2)^2=(3(1)-2)^2\]=1\[rhs=\frac{ n(6n^2-3n-1) }{ 2 }=\frac{ 1(6(1)-3(1)-1) }{ 2}=1\]

Answer 28

thank you so much

Answer 29

therefore LHS=RHS, do you see dat

Answer 30

yes

Answer 31

\[\text{Use mathematical induction to prove the statement is true}\\
\text{for all positive integers n, or }\textit{show why it is false.}\]

Answer 32

The statement is not necessarily true.

Answer 33

the thing is we not done

Answer 34

the inductive hypothesis: for k, a fixed but arbitrary integer with k>=1, assume that results holds when n=k, such that:\[1^2+4^2...(3k-2)^2=\frac{ k(6k^2-3k-1) }{ 2 }\]

Answer 35

do you see that