Show Arithmetic Sequence: V0=4, Vn+1=√Vn2+2n+3

[mtayab1994]

Homework Statement

V0=4

V_{n+1}=\sqrt{V_{n}^{2}+2n+3}

Homework Equations

Show that Un is an arithmetic sequence.

The Attempt at a Solution

I counted Vn and i found that it equals:

V_{n}=\sqrt{(Vn+2)^{2}+2}

what is there to do after this?

 

[SammyS]

Homework Statement

V0=4

V_{n+1}=\sqrt{V_{n}^{2}+2n+3}

Homework Equations

Show that Un is an arithmetic sequence.

The Attempt at a Solution

I counted Vn and i found that it equals:

V_{n}=\sqrt{(Vn+2)^{2}+2}

What is there to do after this?

What is U_n? Is that a typo, or is U_n = (V_n)² ?

 

[mtayab1994]

What is U_n? Is that a typo, or is U_n = (V_n)² ?

No there is no Un at all.

 

[SammyS]

...

Homework Equations

Show that Un is an arithmetic sequence.

That looks like a U_n to me.

BTW: How do you count V_n ?

 

[mtayab1994]

That looks like a U_n to me.

BTW: How do you count V_n ?

Sorry it's show that Vn is arithmetic

 

[Curious3141]

Sorry it's show that Vn is arithmetic

Well, it clearly can't be, because V_0 = 4, V_1 = \sqrt{19} and V_2 = \sqrt{24} and V_2 - V_1 \neq V_1 - V_0 establishing that there is no common difference.

 

[mtayab1994]

sorry V0=1

V0=1 V1=√6 V2=√11 V3=√16


I found that Un=1+√(1+5n)

And i know that arithmetic series are written as Un=Up+nr

so: Up=1 and r=5 therefore you get: Un=1+√(1+5n)

is that all I have to do?

 

[Curious3141]

sorry V0=1

V0=1 V1=√6 V2=√11 V3=√16


I found that Un=1+√(1+5n)

And i know that arithmetic series are written as Un=Up+nr

so: Up=1 and r=5 therefore you get: Un=1+√(1+5n)

is that all I have to do?

What is U_n? You've only defined what V_n is so far.

 

[Curious3141]

By asking what U_n is, I don't mean just quote a formula which you've derived. Please define exactly what U_n is supposed to represent.

It might be better if you reproduced the exact question in its original form, word for word.

 

[mtayab1994]

By asking what U_n is, I don't mean just quote a formula which you've derived. Please define exactly what U_n is supposed to represent.

It might be better if you reproduced the exact question in its original form, word for word.

well the general form of an arithmetic series is :

Un=Up+nr and in my case Un is Vn and Up is V0 and r is 5 so i get:

V_{n}=1+\sqrt{1+5n}

 

[Curious3141]

well the general form of an arithmetic series is :

Un=Up+nr and in my case Un is Vn and Up is V0 and r is 5 so i get:

V_{n}=1+\sqrt{1+5n}

OK, it's all clearer now. Part of the confusion lay in the fact that you had miscalculated the terms for V_n repeatedly.

Forget about trying to fit things into a particular form for now. Let's start by looking at some values of V_n, correctly computed.

Please recalculate V_1, V_2, V_3 and V_4 very carefully, and you'll see a much simpler pattern emerging. We'll take it from there.

 

[Curious3141]

After you do that, you will need to find a closed form expression for V_n (the one you previously derived is clearly wrong), then formally prove it with mathematical induction. Once that's done, it'll become immediately apparent that V_n is the general term of an arithmetic progression (in fact, one of the simplest and most well-known arithmetic progressions).

 

[mtayab1994]

after you do that, you will need to find a closed form expression for v_n (the one you previously derived is clearly wrong), then formally prove it with mathematical induction. Once that's done, it'll become immediately apparent that v_n is the general term of an arithmetic progression (in fact, one of the simplest and most well-known arithmetic progressions).

alright i'll count them right now.

 

[mtayab1994]

i keep getting V0=1 V1=√6 V2=√11 V3=√16

idk what is wrong?

 

[Curious3141]

i keep getting V0=1 V1=√6 V2=√11 V3=√16

idk what is wrong?

V0 = 1 (given)

V1 = sqrt(1 + 2*0 + 3) = sqrt (4) = ?

Once you get V1 wrong, the rest will be wrong too, so restart from here.

You were probably doing sqrt(1+2*1 + 3), but remember the index for the initial term is zero.

 

[mtayab1994]

Wow I didn't pay attention to that sorry:

V0=1 V1=2 V2=3 V3=4 and so one

so the general difference is 1

 

[mtayab1994]

V0 = 1 (given)

V1 = sqrt(1 + 2*0 + 3) = sqrt (4) = ?

Once you get V1 wrong, the rest will be wrong too, so restart from here.

You were probably doing sqrt(1+2*1 + 3), but remember the index for the initial term is zero.

Yes I didn't pay attention to that.

 

[mtayab1994]

Vn in terms of n is Vn=1+n

 

[mtayab1994]

To check I did Vn+1-Vn=2+n-1-2=1

so the common difference is 1.

do i also have to so Vn+2-Vn+1?

 

[Curious3141]

To check I did Vn+1-Vn=2+n-1-2=1

so the common difference is 1.

do i also have to so Vn+2-Vn+1?

No need. You've now got an expression for V_n. You need to prove it.

Two ways.

First is a direct proof, which might proceed like so:

V_{n}^2 = V_{n-1}^2 + 2(n-1) + 3

V_{n-1}^2 = V_{n-2}^2 + 2(n-2) + 3

...

V_1^2 = V_0^2 + 2(0) + 3

then successively substituting the equation below into the one above until one gets:

V_{n}^2 = V_0^2 + 2(\frac{1}{2})(n-1)(n) + 3n

which can be simplified to:

V_{n}^2 = {(n+1)}^2

V_{n} = n+1

Fairly simple. But I would recommend the second method, mathematical induction. Try and do this as an exercise, and post your results here.

 

[mtayab1994]

No need. You've now got an expression for V_n. You need to prove it.

Two ways.

First is a direct proof, which might proceed like so:

V_{n}^2 = V_{n-1}^2 + 2(n-1) + 3

V_{n-1}^2 = V_{n-2}^2 + 2(n-2) + 3

...

V_1^2 = V_0^2 + 2(0) + 3

then successively substituting the equation below into the one above until one gets:

V_{n}^2 = V_0^2 + 2(\frac{1}{2})(n-1)(n) + 3n

which can be simplified to:

V_{n}^2 = {(n+1)}^2

V_{n} = n+1

Fairly simple. But I would recommend the second method, mathematical induction. Try and do this as an exercise, and post your results here.

I posted my mathematical induction on a different thread. I called it Math Series. Please go check it out and tell me if its good.

 

[Curious3141]

I posted my mathematical induction on a different thread. I called it Math Series. Please go check it out and tell me if its good.

Just to round off your answer, state that V_n is an arithmetic progression (AP) because V_{n+1} - V_{n} = 1, which is a constant (the common difference). Hence the AP has first term 1 and common difference 1. In fact, the sequence comprises the natural numbers.

I took a brief look at the other thread - it's an unrelated question. Afraid I can't look at this now as it's past 1 am my local time and I need to sleep, so someone else may step in and help you. However to do the induction for this problem, you need to establish the result for V_1 (show that what you work out from the recursive square root formula, i.e. \sqrt{1 + 2(0) + 3} is equal to the closed form formula, i.e. 1+1, which is trivial, then prove that assuming the result for a particular V_k leads to the result for the next term V_{k+1}. This is just simple algebra.

Having given you this hint, I'll turn in now. Good luck.

 

[mtayab1994]

Just to round off your answer, state that V_n is an arithmetic progression (AP) because V_{n+1} - V_{n} = 1, which is a constant (the common difference). Hence the AP has first term 1 and common difference 1. In fact, the sequence comprises the natural numbers.

I took a brief look at the other thread - it's an unrelated question. Afraid I can't look at this now as it's past 1 am my local time and I need to sleep, so someone else may step in and help you. However to do the induction for this problem, you need to establish the result for V_1 (show that what you work out from the recursive square root formula, i.e. \sqrt{1 + 2(0) + 3} is equal to the closed form formula, i.e. 1+1, which is trivial, then prove that assuming the result for a particular V_k leads to the result for the next term V_{k+1}. This is just simple algebra.

Having given you this hint, I'll turn in now. Good luck.

Thank you very very very much for your help.

 

[Curious3141]

Thank you very very very much for your help.

You're welcome. You're also welcome to post the crucial inductive step of your proof for this problem here, and I (or someone else) can check it. Unless you're sure of it, then it's OK.

 

[mtayab1994]

You're welcome. You're also welcome to post the crucial inductive step of your proof for this problem here, and I (or someone else) can check it. Unless you're sure of it, then it's OK.

Yea I'm sure of it thank you anyway.