Question

Please Help!!

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))/6

Answer 1

@SolomonZelman

Answer 2

@zzr0ck3r

Answer 3

yes

Answer 4

so you want to prove \(\sum_{i=1}^n4i(4i+1)=\frac{4(4n+1)(8n+7)}{6}\)

Answer 5

ill write it out and upload one sec

Answer 6

ok, thank you :)

Answer 7

I dont think it does

Answer 8

well it also says "or show why it is false"

Answer 9

4*6+5*7+6*8 = 107
4(4*3+1)(8*3+7)/6 = 806/3

Answer 10

so I looked for n = 3

Answer 11

why does n=3

Answer 12

I think you are confused on what we are doing

you have a statement that says something is true for all n, I showed that it is not true for n = 3, thus it is NOT the case that it is true for all n

Answer 13

so wouldn't it be 4*6+5*7+6*8+4(3)(4*3+2)=(4(4*3+1)(8*3+7))/6

Answer 14

something is messed up

for the sequence 4n(4n+2) we get

4*6, 8*10, 12*14

reread the question and make sure you posted it correctly

Answer 15

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

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

Answer 16

do you see why this is messed up?

Answer 17

4*6+5*7+6*8 ... does not come from 4n(4n+2)

Answer 18

why?

Answer 19

let n=1 and you get 4*(4+2)=4*6 as this is good so far.
let n = 2 and we get 4*2(4*2+1)=8*9 NOOOOO

Answer 20

ohhh ok

Answer 21

so is that what i would write?

Answer 22

your sequence says we shuold get 4*6+5*7
not 4*6+8*10

Answer 23

I would write that the question does not make sense.

Answer 24

there is a typo

Answer 25

can you help me with another one?

Answer 26

I am not sure if we shuold consider the sequence 4*6+5*7+6*8 or 4*6+8*10+12*14...

Answer 27

sure

Answer 28

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

Answer 29

@zzr0ck3r

Answer 30

hey @wio can you help with this??

Answer 31

lol I said I would

Answer 32

no no i know you are but hes just sitting there...

Answer 33

watching...

Answer 34

again the question makes no sense....nothing squared is 12

Answer 35

where are you getting these?

Answer 36

my math assignment from flvs

Answer 37

he sequence you showed shuold say

12+42+72+......+n^2-n

Answer 38

starting with n = 4

Answer 39

but it doesn't >.<

Answer 40

@wio do you have any idea whats going on here?

Answer 41

Is it the same question?

Answer 42

they are not making sense

Answer 43

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

Answer 44

they are giving the wrong definition for the sequences (3n-2)^2 for this one.

Answer 45

and something strange on the last one

Answer 46

shuold say
12 + 42 + 72 + ... + n^2-n =
starting at n=4

Answer 47

@zzr0ck3r @wio

Answer 48

For the first problem, the nth term they have given is wrong.

The nth term should be: (n+3)(n+5)

n = 1: 4*5
n = 2: 5*7
n = 3: 6*8

Answer 49

It is not clear whether they actually want the sum of the series or just to prove whether the nth term is correct or not.

The nth term is wrong on the first one because if you put n = 2 you get 8*10 which is not the second term in the series.

The nth term on the second one is correct because if you put n = 1, you get 1^2, n = 2 gives 4^2, n = 3 gives 7^2.

Answer 50

they just want to know if it is true or false and how i got to that conclusion but i didn't know how to go about it.

Answer 51

@aum

Answer 52

I see, so \(5\times 7\) is not a term in the sequence \(4n(4n+2)\).

Answer 53

so what does 4n(4n+1)(8n+7)/6 have to do with anything

Answer 54

Would the sequence be \((n+3)(n+5)\)

Answer 55

i have no clue, i am so confused

Answer 56

ok ninja this is how it would normally work

1+3+5+.....+(2n-1)=n^2

what that means is 1+3+5+7+9..... is a sequence that is given by the rule 2n-1
i.e. if we let n=1 we get 1, if we let n=2 we get 3, if we let n=3 we get 5.

in the two series you gave us, the rule does not match the sequence before the rule

Answer 57

@ninjasandtigers

Answer 58

uh ok thank you. @zzr0ck3r