a(n)=4a(n-1) + 4 if n is odd
a(n)=4a(n-1) - 4 if n is even
with a(0)=0
as i think you will find idt useful.
e.g.
a(1) = a(0)*4 + 4 = 0*4 + 4 = 4
a(2) = a(1) *4 - 4 = 4*4 - 4 = 12 n is even
and so on...
I'm to find an explicit formula for this sequence:
a_1 = 0
a_2 = 4
a_3 = 12
a_4 = 52
a_5 = 204
a_6 = 820
...
(ie "add 1 and muliplty by 4; subtract 1 and multiply by 4; add 1 and muliplty by 4; subtract 1 and multiply by 4 etc etc")
But can an explicit formula for a_n in terms of n (I don't want a "recurrent" formula based on a_(n-1), obviously) be found?
a(n)=4a(n-1) + 4 if n is odd
a(n)=4a(n-1) - 4 if n is even
with a(0)=0
as i think you will find idt useful.
e.g.
a(1) = a(0)*4 + 4 = 0*4 + 4 = 4
a(2) = a(1) *4 - 4 = 4*4 - 4 = 12 n is even
and so on...
nth term of series is as follows
a(n)=(-1)^0*2^(2*n) + (-1)^1 * 2^(2*n - 2) +....while 2^0 not comes if it comes leave it
e.g.
a(0)=(-1)^0 * 2^(2*0) = 0
a(1) = (-1)^0 * 2^(2*1) + (-1)^1 * 2^(2 - 2) but second will be igonered. = 4
similarly:
it follows for all other elements of the series.......