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

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 $3/2+21/4+63/8+177/16....$?

• Where has the 177 come from? It's $3 \times 59$, but that doesn't seem to have any link with the previous things. – Patrick Stevens Oct 23 '15 at 14:17
• I know right! I had this in my problem set i did not know what to do.. so asked this question looking for hints. Is the question wrong ? – user283172 Oct 23 '15 at 14:20

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

• I think the question is wrong but thanks anyway! – user283172 Oct 23 '15 at 14:50
• I personally don't see any pattern in the formulas you wrote. – Wojowu Oct 23 '15 at 14:50
• Yea I think as OP mentioned, the question could be wrong – Nicholas Oct 23 '15 at 14:51

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

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