mathematics

[Silent Hunter]

hey people do you know any specific and easy method of finding formulae for 'n' th term of any sequence..........??/

 

[CaptainDanger]

Yeah....

Here it is

For linear sequence
a+(n-1) d1

for
Quadratic sequence
a+(n-1) d1 + 1/2 (n-1) (n-2) d2


a is the first term of sequence.
d1 is the first difference
d2 is the second difference.

Put accordingly, solve and you will arrive at the n-th term formula...

 

[rafae]

I also have another formula.
For a linear difference, the formula goes as such,
T=An + B

A=Difference.
'n'=the number of which you are solving
Tn= the number of 'n'

I'll explain it look:

n=1, 2, 3, 5, 6
= 2, 3, 4, 5, 6

So here it would be, take any number from 'n'
If i take 'n' as 1, it would be
Tn= An + B
A=difference so A=1
If i take the number 1, then the number value of 1 is 2 so Tn is 2
2= (1)(1) + B
B=1
Formula becomes n+1


I also have one for quadratic equations, its a bit different to CaptainDanger's, i'll post it in a bit.

 

[CaptainDanger]

Thanks .... Didn't know that....

 

[rafae]

Okay, here's the one for quadratic equation.
This is a bit more complex.
An² + Bn + C= Tn

The principal is the same except for a few changes.

A= difference, but this time, the difference will be divided by two. It will ALWAYS be divided by two.
B & C are both unknowns, and therefore you will obtain two equations. You will have to solve them simultaneously. If you have any problems with solving the equation, then just private message me.

 

[insomniac]

Okay, here's the one for quadratic equation.
This is a bit more complex.
An² + Bn + C= Tn

The principal is the same except for a few changes.

A= difference, but this time, the difference will be divided by two. It will ALWAYS be divided by two.
B & C are both unknowns, and therefore you will obtain two equations. You will have to solve them simultaneously. If you have any problems with solving the equation, then just private message me.

well can you explain it further by giving an instance .. Thank you!

 

[rafae]

Here you go insomniac.
Okay, here you go.

Well if this is the question:

number: [1,] [ 2,] [3,] [4,] [5]
Tn=.......[4,] [7,] [12,] [19,] [28]

Okay this gets a bit complex. You have to find out the second difference.
So:
4, 7, 12, 19, 28
(3...5...7...9)
(....2...2...2)

What I have written in brackets are the differences between each 'Tn'
So you have a difference of 2 in the second difference.
In quadratic equations, you always divide by 2.
So A=2/2, =1

First equation:
An² + Bn + C= Tn
I'm taking the first number now.
1(1)² + B(1) + C=4
1+B+C=4

(i)B+C=3 (Equation one)


An² + Bn + C= Tn
1(2)² + B(2) + C = 7
4 + 2B + C = 7
(ii)2B + C= 3 (Equation two)

Now, simultaneously

B=3-C (from equation one)

2(3-C) + C= 3
6- 2C + C = 3
6 -C = 3
C=3
B=0
Just solve it simultaneously.
Now put it into the equation
n²+3, there you go.

 

[insomniac]

umm a lil confused ... a=? ... is a=1 bu how come when the first term of the sequence is 4??

 

[CaptainDanger]

You have to put the Tn number there... 4 is the first number of sequence so put 1 there....



See mine if it helps...

Find an expression in terms of n for this sequence...

3, 6, 10, 15, 21, ___ , ___


a + (n-1)d1 + 1/2(n-1)(n-2)d2

a is the 1st term of the sequence
d1 is the 1st difference
d2 is the second difference!

==> 3+(n-1)3 + 1/2 (n-1)(n-2)1
= (3 + 3n - 3) + 1/2 (n^2 - 2n - n + 2)
= (3n) + (n^2 - 3n + 2) /2
= (6n + n^2 - 3n + 2)/2
= (n^2 + 3n + 2)/2
= (n^2 + 2n + n + 2)/2
= (n+1)(n+2)/2

 

[insomniac]

You have to put the Tn number there... 4 is the first number of sequence so put 1 there....



See mine if it helps...

Find an expression in terms of n for this sequence...

3, 6, 10, 15, 21, ___ , ___


a + (n-1)d1 + 1/2(n-1)(n-2)d2

a is the 1st term of the sequence
d1 is the 1st difference
d2 is the second difference!

==> 3+(n-1)3 + 1/2 (n-1)(n-2)1
= (3 + 3n - 3) + 1/2 (n^2 - 2n - n + 2)
= (3n) + (n^2 - 3n + 2) /2
= (6n + n^2 - 3n + 2)/2
= (n^2 + 3n + 2)/2
= (n^2 + 2n + n + 2)/2
= (n+1)(n+2)/2

alright with this method can you please help me in solving this question
0,3,8,15.... are the first 4 terms of a sequence.. write down an expression in terms of n, for the nth term ...

 

[insomniac]

You have to put the Tn number there... 4 is the first number of sequence so put 1 there....



See mine if it helps...

Find an expression in terms of n for this sequence...

3, 6, 10, 15, 21, ___ , ___


a + (n-1)d1 + 1/2(n-1)(n-2)d2

a is the 1st term of the sequence
d1 is the 1st difference
d2 is the second difference!

==> 3+(n-1)3 + 1/2 (n-1)(n-2)1
= (3 + 3n - 3) + 1/2 (n^2 - 2n - n + 2)
= (3n) + (n^2 - 3n + 2) /2
= (6n + n^2 - 3n + 2)/2
= (n^2 + 3n + 2)/2
= (n^2 + 2n + n + 2)/2
= (n+1)(n+2)/2

and yeah if we are supposed to put the Tn value for'a' then its always gonna b 1 bcoz the tn is always 1 .. that means there is no need for a in the previous formula.. (hell confusion)

 

[CaptainDanger]

alright with this method can you please help me in solving this question
0,3,8,15.... are the first 4 terms of a sequence.. write down an expression in terms of n, for the nth term ...

Yes
0,3,8,15

See the first difference is 3

and the second 2

a + (n-1) d1 +1 /2 (n-1) (n-2) d2

a is the first number of sequence again

==>
0+ (n-1) 3 + 1/2 (n-1) (n-2) 2
3n-3 + 1/2 (n^2 - 3n +2) 2
3n-3 + 1/2 (2n^2 - 6n + 4)
3n-3 + n^2 -3n + 2
n^2-1 becomes the formula....

 

[CaptainDanger]

and yeah if we are supposed to put the Tn value for'a' then its always gonna b 1 bcoz the tn is always 1 .. that means there is no need for a in the previous formula.. (hell confusion)

No Tn is not always one.... See how I solved it.... Use that formula...

 

[insomniac]

now it seems really good ... thanks .. =)

 

[CaptainDanger]

You are most welcome...

 

[insomniac]

 

[Al. Ahmad]

I also have another formula.
For a linear difference, the formula goes as such,
T=An + B

A=Difference.
'n'=the number of which you are solving
Tn= the number of 'n'

I'll explain it look:

n=1, 2, 3, 5, 6
= 2, 3, 4, 5, 6

So here it would be, take any number from 'n'
If i take 'n' as 1, it would be
Tn= An + B
A=difference so A=1
If i take the number 1, then the number value of 1 is 2 so Tn is 2
2= (1)(1) + B
B=1
Formula becomes n+1


I also have one for quadratic equations, its a bit different to CaptainDanger's, i'll post it in a bit.

B=1, but how. didnt get it..:/

 

[talal096]

i need help in vectors??

 

[Momin]

You have to put the Tn number there... 4 is the first number of sequence so put 1 there....



See mine if it helps...

Find an expression in terms of n for this sequence...

3, 6, 10, 15, 21, ___ , ___


a + (n-1)d1 + 1/2(n-1)(n-2)d2

a is the 1st term of the sequence
d1 is the 1st difference
d2 is the second difference!

==> 3+(n-1)3 + 1/2 (n-1)(n-2)1
= (3 + 3n - 3) + 1/2 (n^2 - 2n - n + 2)
= (3n) + (n^2 - 3n + 2) /2
= (6n + n^2 - 3n + 2)/2
= (n^2 + 3n + 2)/2
= (n^2 + 2n + n + 2)/2
= (n+1)(n+2)/2

This helped in my paper and ironically same sequence came