Multiplication of summations The length of the following equation depends on the value of the variable n. How can I write this equation? How can I reformulate it?
$$f(n)=x+y\cdot\sum_{a=1}^{n}\left(x+y\cdot\sum_{b=1}^{n-1}\left(x+y\cdot\sum_{c=1}^{n-2}\left(x+y\cdot\sum_{d=1}^{n-3}....\right)\right)\right)$$
Moreover I'd need that the sum $\sum_{1}^{1}$ equals 0!
for example:
$$f(3)=x+y\cdot\sum_{a=1}^{3}\left(x+y\cdot\sum_{b=1}^{2}\left(x+y\cdot\sum_{c=1}^{1}\left(x+y\cdot0\right)\right)\right)$$
UPDATE
Re-written differently I think it gives. But still, the length of my equation depends on the value of n
$$f(n)=x+y \cdot (n \cdot x + n \cdot y \cdot ((n-1)\cdot x + (n-1) \cdot y \cdot ((n-2)\cdot x + (n-2)\cdot y \cdot (....)) ))$$
for example:
$$f(3)=x+y \cdot (3 \cdot x + 3 \cdot y \cdot ((2)\cdot x + (2) \cdot y \cdot ((1)\cdot x + (1)\cdot y \cdot 0)))$$
Thank you!

Given that summing $m$ times the same thing amounts tom multiplying it by $m$, you get
$$
  f(n)=x+yn(x+y(n-1)(x+y(n-2)(\cdots(x+y.1(x+0y))\ldots)))
$$
which resembles a polynomial in $y$ expressed by a Horner scheme; you get
$$
\begin{align}
  f(n)&=x+xny+xn(n-1)y+n(n-1)(n-2)y^3x+\cdots+n(n-1)\ldots2xy^{n-1}+n!xy^n
\\ &=x\sum_{i=0}^nn^{\underline i}y^i
\end{align}
$$
where $n^{\underline i}$ denotes the falling factorial power.
