Math - Sequences/Series

[Flikerator]

Im doing a math assignment right now, and the question is this;

code:

If t[n] = ((t[n-1])**2) + 4n-1 and t[1]=1, find terms t[2] through t[5].

Where the values in [] are subscripts.

So if im doing t[2];

t[2] = ((t[2-1])**2) +4(2)-1
= ((t[1])**2) + 7

Would this be 8? Because t[1] = 1?

In which case, this would be recrusive, which I can do. I just need to know if its right so I don't mess everything up ><

Edit: Also, wouldn't the "**2" always result in 1??

[Cervantes]

Standard notation for subscripts would look like "t_n" where n is a subscript for t.

That's right. t_2 would be 7.

No, squaring t_n-1 does not always result in 1. What if n = 3? Then n-1 = 2. t_2 is not 1, it's 7. So 7 ** 2 is 49, not 1.

[Mazer]

Looks ok to me. Except for the **2 always giving one. When you're trying to find t[3], you'll be subbing in t[2], so it'd be 8**2 that time.

[Flikerator]

Where {S} = Sigma Notation (ie upper case greek letter Sigma);

EDIT: t[n] = n**4

code:

n {s}i**3 =[ (n(n+1))**2 ] / 4     <--- Its all divided by 4 i=1

Any place to START on this question would be great. Its the fourth Summation question, and unlike the others, I don't know what to do. Thanks for your time..

[Flikerator]

Cervantes wrote:

Standard notation for subscripts would look like "t_n" where n is a subscript for t.

That's right. t_2 would be 7.

No, squaring t_n-1 does not always result in 1. What if n = 3? Then n-1 = 2. t_2 is not 1, it's 7. So 7 ** 2 is 49, not 1.

t_n = (t_n-1)**2 + 4n - 1

t_1 = 1

t_2= (t_2-1)**2 + 4(2)-1
=(t_1)**2 + (8-1)
=(1)**2 + (7)
= 8

t_3 = (t_3-1)**2 + 4(3)-1
=(t_2)**2 + (12-1)
=(t_2-1)**2 + 11???

t_2 = t(2) = t_2-1

Wouldn't that be the case? because the subscript for t results in the t function with the "x" less then 1.

f(x) = f(x-1)
Resulting in;
f(x-1) = f(x-1-1) = f(x-2)

Or am I just HIGHLY confused? Thanks for all the help, I posted that over the course of like an hour doing questions so I didn't read your responses.

EDIT: My explanation is a little hard to follow (I think anyways). So I tried to clear it up with the f(x) thing...

[Cervantes]

What's the question? Is it asking you to prove that?

To derive it, you will need to know the formulas for the sum of the natural numbers and the sum of the squares of the natural numbers.

To prove it, you can use mathematical induction.

[Flikerator]

Cervantes wrote:

What's the question? Is it asking you to prove that?

To derive it, you will need to know the formulas for the sum of the natural numbers and the sum of the squares of the natural numbers.

To prove it, you can use mathematical induction.

Yes its asking to prove that;

code:

n {s}i**3 =[ (n(n+1))**2 ] / 4 i=1

Im brushing up on my induction right now actually, we've done a week of it so I should be able to figure something out. Thanks for the help! ^_^

[Cervantes]

Flikerator wrote:

t_3 = (t_3-1)**2 + 4(3)-1
=(t_2)**2 + (12-1)
=(t_2-1)**2 + 11???

t_2 = t(2) = t_2-1

...

Or am I just HIGHLY confused? Thanks for all the help, I posted that over the course of like an hour doing questions so I didn't read your responses.

You already know the value of t_2. It's 8.
If you really want to though, you could sub in further (you subbed in wrong):
t_3 = t_2 ** 2 + 11
t_3 = (t_1 ** 2 + 4(2) - 1) ** 2 + 11
t_3 = (1 ** 2 + 7) ** 2 + 11
t_3 = 75

[Flikerator]

I don't need you to do these for me, I wouldn't learn anything. Just some help into what to do for them...Im really lost and would really appreciate any help..

How do I solve the following;

(18)
(5 )

is it;

n!
--------
r!(n-r)!

?

code:

(2x-3)**10

Any easy way to expand that? Binomial theorem with Pascals triangle comes to mind, but i can't remember...

a)Give a general term
b)Find the middle term.
c)What is the coefficient of the term with x**3?

The next question;

code:

(x+4)(3x-5)**7

a)What is the coefficient of the term with x**8?
b)Find the middle term.


Along the same lines are these;

code:

(a-3b)**5 (2x - (y/3))**7 (3w-(1/w))**6

Again I don't really want you to solve them for me, just how to do them.

[Flikerator]

Cervantes wrote:

Flikerator wrote:

t_3 = (t_3-1)**2 + 4(3)-1
=(t_2)**2 + (12-1)
=(t_2-1)**2 + 11???

t_2 = t(2) = t_2-1

...

Or am I just HIGHLY confused? Thanks for all the help, I posted that over the course of like an hour doing questions so I didn't read your responses.

You already know the value of t_2. It's 8.
If you really want to though, you could sub in further (you subbed in wrong):
t_3 = t_2 ** 2 + 11
t_3 = (t_1 ** 2 + 4(2) - 1) ** 2 + 11
t_3 = (1 ** 2 + 7) ** 2 + 11
t_3 = 75

Oh damn! XD Thanks a lot. Can't believe I missed that, even after you both explained it...thanks again!

[Flikerator]

(a**5)
- (15a**4b)
+ (30a**3b**2)
- (30a**2b**3)
+ (15ab**4)
- (b**5)

I know its hard to read, I put it line by line to make it possibly a tad easier to read them. Heres it on one line;

(a**5) - (15a**4b) + (30a**3b**2) - (30a**2b**3) + (15ab**4) - (b**5)

I know its a pain in the ass, but could someone check to make sure this is correct? Its the easiest, and I really don't want to continue if im doing it wrong;

(a-3b)**5

EDIT:Nevermind, got the real answer...yey for learning! ^_^

[Flikerator]

code:

k(k + 1) + 2(k + 1) --------------------          2 How does it become; Code: (k + 1)(k + 2) --------------        2

Im very tired...School + 8 or so hours of math. I have everything done except question 4 and 5... Im doing 4 last. W/e

I have the whole induction thing down (mostly) except for that. How does that work? Its not even a problem with induction, just the fraction or what have you...tired though...

If anyone could help me, Id be very grateful. Im almost done my math, huzzah.

EDIT: WOOT I figured it out!

[TheOneTrueGod]

common factoring. 2a + 3a = (2+3)a = 5a, correct? Apply this to your problem (Use some substitution), and you'll see

[Cervantes]

For the binomial expansion, you can just try subbing in two values for a and b and seeing if it works. If it does, that's a good indicator that you did it right.

Where are you doing all thi? I thought you were in grade 10. I know my geometry and discrete class is only covering this now!

[Flikerator]

Yah im grade 11, taking a grade 12 math course (Discrete Mathematics). Its only offered this year so I had to take it.

Its actually not that hard, I've just been slacking all year not doing any of the work. Now that im doing it its not so bad. Can't turn back time you know?

I couldn't see the common factoring last night. I see it now though. I blame being tired. We were given an extra day to do the assignment, so after work I'll be working on it.

Thanks for the explanation TheOneTrueGod, I forgot all about how to common factor ^_^ (Yah I must have been really tired )

Still can't figure out that last Sigma Notation question though. Im working on it though =P

Edit: Only have the sigma notation and if I can manage to factor;

code:

(k+1) [4k**2 + 25k + 28]

I think I did something wrong. The closest I can get is;

(2k + 7) (2k + 4)

But that gets 23k instead of 25k.

EDIT: Always copy the question out properly XDD