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.