0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Where has the 177 come from? It's $3 \times 59$, but that doesn't seem to have any link with the previous things. $\endgroup$ – Patrick Stevens Oct 23 '15 at 14:17
  • $\begingroup$ 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 ? $\endgroup$ – user283172 Oct 23 '15 at 14:20
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I think the question is wrong but thanks anyway! $\endgroup$ – user283172 Oct 23 '15 at 14:50
  • 2
    $\begingroup$ I personally don't see any pattern in the formulas you wrote. $\endgroup$ – Wojowu Oct 23 '15 at 14:50
  • $\begingroup$ Yea I think as OP mentioned, the question could be wrong $\endgroup$ – Nicholas Oct 23 '15 at 14:51
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.