Closed form for $f_k(y)$

Question

The question is quite simple. Given $$\sum_{k=0}^{n-1}(x+y)^k$$ We can re-write it in terms as a polynomial in $x$, with coefficients being polynomials in $y$, i.e $$\sum_{k=0}^{n-1}(x+y)^k = \sum_{k=0}^{n-1}f_k(y)x^k$$ I attempted to re-write it as $$\sum_{k=0}^{n-1}\sum_{r=0}^{k}{k \choose r}x^{k-r}y^r$$ $$ = \sum_{k=0}^{n-1}\left(\sum_{r=0}^{k}{k \choose r}\left(\frac{y}{x}\right)^r\right)x^k$$ I tried equating but came back to the original form. Any help/answer would be appreciated

For example: $n= 3$

$$f_0(y) = 1+y+y^2$$ $$f_1(y) = 1+2y$$ $$f_2(y) = 1$$

so we have $(1+y+y^2)x^0+(1+2y)x+(1)x^2$

Answer 1

You should learn how to do these yourself. $$ \sum_{k=0}^{n-1}(x+y)^k =\sum_{k=0}^{n-1} \sum_{j=0}^k \binom{k}{j} x^j y^{k-j} . $$ Now, for each $j$ collect all the terms with $x^j$. (Reverse the order of summation.) $$ \sum_{k=0}^{n-1} \sum_{j=0}^k \binom{k}{j} x^j y^{k-j}= \sum_{j=0}^{n-1} \sum_{k=j}^{n-1}\binom{k}{j} x^j y^{k-j} . $$ The condition $j \le k$ in the first one yields the condition $k \ge j$ in the second one.
Then factor out from the inner sum things that depend only on $j$ $$ \sum_{j=0}^{n-1} \left(\sum_{k=j}^{n-1}\binom{k}{j} y^{k-j}\right)x^j . $$ In the required notation, $$ f_j(y) = \sum_{k=j}^{n-1}\binom{k}{j} y^{k-j} , \\ f_k(y)= \sum_{i=k}^{n-1}\binom{i}{k} y^{i-k} = \sum_{j=0}^{n-1-k}\binom{k+j}{k} y^{j} . $$ We have to keep track of re-named dummy variables.

Answer 2

It is convenient to use the coefficient of operator $[x^q]$ to denote the coefficient of $x^q$ of a series. We obtain for $0\leq q\leq n-1$:

\begin{align*} \color{blue}{f_q(x)}&=[x^q]\sum_{k=0}^{n-1}(x+y)^k=\sum_{k=q}^{n-1}[x^q](x+y)^k\tag{1}\\ &=\sum_{k=q}^{n-1}\binom{k}{q}y^{k-q}\tag{2}\\ &\,\,\color{blue}{=\sum_{k=0}^{n-1-q}\binom{k+q}{q}y^k}\tag{3} \end{align*}

Comment:

• In (1) we use the linearity of the coefficient of operator. We also start the with index $k=q$, since powers of $x$ with power less than $q$ do not contribute.

• In (2) we select from the binomial the coefficient of $x^q$.

• In (3) we shift the index to start with $k=0$.