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Maths Number Patterns

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How do we find out the equation for the nth term? I never get that right and as result mess up the entire ques. Any help would be appreciated
 
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For a linear difference, the formula goes as such,
T(n)=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
Tn=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 since the difference between 2 or 3 is 1, 4 or 5 is 1, 5 or 6 is 1, then we will take 1.
If i take 'n' as 1, then the Tn of 1 is (2) so Tn = 2
2= (1)(1) + B
B=1
So you have the value of B now which is 1.
Now put it into the equation.
Tn= n +1
That is your formula.
I'll post quadratic in a bit.
 
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Mine are a little different from Rafae

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...

EXAMPLE


QUADRATIC
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....
 
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No.... The second difference here ummm:

See

0,3,8,15

1st time when you take for all the four it comes 3,5,7

Now again take the difference so it becomes 2, 2.... Its the 2nd time difference... You have to put that there in d2... :)
 
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