Formula for the $n$th term of $3/2+21/4+63/8+177/16....$

Question

Any idea about how to go about doing this any hints on how to get strated with this ?
I should find a closed expressing in terms of "n" for the first n terms of of this sequence $3/2+21/4+63/8+177/16....$?
Thanks In Advance!

Answer 1

Just an idea for the numerator.

$3+18=21$

$21+(18+24)=63$

$63+(18+3(24))=177$

Not sure if it holds past $177/16$ though

Answer 2

It should be $63 + 18 + 4\times 24 = 177$

Answer 3

Just another ideia for the numerator:

3*1*2^0+3^1 = 003+003 = 006

3*2*2^1+3^1 = 012+009 = 021

3*3*2^2+3^1 = 036+027 = 063

3*4*2^3+3^1 = 096+081 = 177

3*5*2^4+3^1 = 240+243 = 483

So the series can be written:

S = 6/2+21/4+63/8+177/16+....-3/2