how to find a100 ?

Question

If a1 = 1 and

for every positive integer n, the eqn below holds
http://rogercortesi.com/eqn/tempimagedir/eqn1830.png

then a100 equals ?

My soln :
---------------
we have
a1=1

From the eqn putting value of a1 i got

a2=5

further putting value of a1,a2 i got

a3=21

but i need to find a100 .

How ?

Answer 1

You need to solve the recurrence equation
a_(n+1) - 3a_(n) = 4n - 2 /here _(n) is a subscript/
According the general theory the solution is given by
a_(n) = C*3^n + bn + c /b, c, C = const; b and c to be determined by substitution, we obtain b = -2, c = 0, the other constant C is determined by the condition a_(1)=1, so C=1/

Finally a_(n) = 3^n - 2n, so a_(100) = 3^100 - 200

Answer 2

1, 5, 21, 73, 233 .....
4,16, 52, 160 <-- 1st differences
12, 36 , 108 <-- 2nd differences
The 2nd differences are 12, 12*3, 12* 3^2 , 12* 3(n-2)
So 12*3^98 = 98th 2nd difference.
You can now work backward to get 99 1st difference and from there to a100.

Answer 3

By observation, a_n = 3^n - 2n. Therefore a_100 = 3^100 - 200.

You can prove this by induction.