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!


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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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