1
$\begingroup$

Let the numbers $c_i$ be defined by the power series identity $(1+ x + x^2+ ... + x^n)/(1-x)^{p-1} = 1 + c_1x + c_2x^2+... $. Show that $c_i \equiv 0 \bmod{p}$ for all $i \geq 1$

Any help on starting out, please? Thank u!

$\endgroup$
3
$\begingroup$

Hint: Since for all suitable $x$ we have $1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$, we are looking at $$\frac{1-x^{n+1}}{(1-x)^p}.$$ Now write down the power series expansion of $\frac{1}{(1-x)^p}$.

Remark: Note however that in general we do not have $c_i\equiv 0\pmod{p}$. Try for example $\frac{1+x}{(1-x)^2}$ (so $n=1$, $p=3$). But for suitable values of $n$, the result holds.

$\endgroup$
  • $\begingroup$ oh damn! sorry, i forgot to write down the work I've already done. sorry, worked on this three days ago and forgot... $\endgroup$ – phoenix Mar 6 '14 at 4:33
  • 1
    $\begingroup$ If you had reached this far, I can delete. It really puts one quite close. $\endgroup$ – André Nicolas Mar 6 '14 at 4:35
  • $\begingroup$ yes, i had done that part. sorry, my internet is being very inconsistent T_T i'm having trouble editing and posting $\endgroup$ – phoenix Mar 6 '14 at 5:00
  • $\begingroup$ $1+px+(p^2+p)x^2/2+(p^3+3p^2+2p)x^3/6+...$ right? $\endgroup$ – phoenix Mar 6 '14 at 5:03
  • $\begingroup$ actually i see now! i had done a calculation mistake and hadnt gotten the exact expression as above. thats why it didnt made sense! $\endgroup$ – phoenix Mar 6 '14 at 5:05

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.