Question

\[n \ge 2,\]\[\sum_{k=0}^{n-1}(x^{2n})^k=g_{n}(x)\sum_{k=0}^{n-1}x^k+n\]\[Find\ \ g_{n}(x)\ \ which\ is\ a\ polynomial \ with \ real \ coefficients.\]

Answer 1

well both are GPs (ones under summation sign)
i hope this'll help !

Answer 2

<ding> ?

Answer 3

i can't get the polynomial.

Answer 4

leme try..
we have
((x^2(n^2)) -1)/(x^2n -1) = gx (x^n -1)/(x-1) + n
i'd guess just solve for gx..not pretty sure if it can be simplified further..

Answer 5

but i can't get the polynomial form.
i just get\[\frac{(1-x) (1-(x^{2n})^n+n (-1+x^{2n}))}{(-1+x^n) (-1+x^{2n})}\]

Answer 6

ohh i see..i'll tell you in some time if i can help or not :P

Answer 7

what did you do to n??

Answer 8

this?\[\frac{(1-x) (1-(x^{2n})^n+n (-1+x^{2n}))}{(-1+x^n) (-1+x^{2n})}\]

Answer 9

lol @experimentX ,,buddy dont worry,,he has done it correctly..

Answer 10

lol ... prof, can you show the trick a bit??

Answer 11

i think it is not correct

Answer 12

I am still verifying it

Answer 13

May be, stay tuned.

Answer 14

Here it is
\[
(x-1) \left(x^n+1\right) \sum
_{k=1}^n (n-k) x^{2 (k-1) n}
\]

Answer 15

correct :)

Answer 16

how to get it?

Answer 17

\[
f[x_]=\sum_{i=0}^n x^i \\

f(x^{2n})=\left((x-1) \left(x^n+1\right) \sum
_{k=1}^n (n-k) x^{2 (k-1) n}\right) f(x) + n
\]

Answer 18

If you read my proof of the related problem I posted, you can realize it as
\[
(x-1) h(x^n)
\]

Answer 19

yes

Answer 20

Oh ... i guess i got it too
\[ x^{2 (k-2) n}\]