Summation of recursive formula

Question

In modelling a certain problem I find the following sum that I am trying to find a formula for $f_1+f_2+f_3...+f_n$ where \begin{align} &f_1=t_1+(a+y(t_1-t_0))k\\ &f_2=t_1+\Bigg(a+2y(t_1-t_0)+\bigg(\Big(a+y(t_1-t_0)\Big)k-ak\bigg)y\Bigg)k\\ &f_3=t_1+\Bigg(a+3y(t_1-t_0)+\bigg(\Big(a+y(t_1-t_0)\Big)k-ak\bigg)y \\ &+ \bigg(\bigg(a+2y(t_1-t_0)+\Big(\big(a+y(t_1-t_0)\big)k-ak\Big)y\bigg)k-ak\bigg)y\Bigg)k\\ &.\\ &.\\ &.\\ \end{align} Notice that the equation of $f_3$ takes up two lines. I found that $f_n$ is given by the recursive formula \begin{align} f_n=t_1+k\bigg( a+ny(t_1-t_0)+y\sum_{i=1}^{n-1} (f_i-t_1-ak) \bigg),\;\;\;\;\; f_1= t_1+k(a+y(t_1-t_0)) \end{align} I know it is very messy and probably very hard to find a formula for the sum or even a non-recursive formula for $f_n$. The pattern is that you add $t_1$ and then some factor multiplied by $k$. Now this factor gets messy and very large quickly. This factor is given by $a+ny(t_1-t_0)$ and then adding each preceding factor minus $ak$ multiplied by $y$. I know its quite a nasty summation, but I will try my luck here!

Answer 1

We can simplify your formula quite a bit, but it doesn't result in a recurrence that I know how to solve. We have $$\begin{align} f_n&=t_1+k\left( a+ny(t_1-t_0)+y\sum_{i=1}^{n-1} (f_i-t_1-ak) \right)\\ &=t_1+ka+kny(t_1-t_0)+ky\sum_{i=1}^{n-1}f_i-ky(n-1)(t_1+ak)\\ &=t_1+ka+n(ky)(t_1-t_0)-n(ky)(t_1+ak)+ky(t_1+ak)+ky\sum_{i=1}^{n-1}f_i\\ &=\alpha+n\beta+ky\sum_{i=1}^{n-1}f_i \end{align}$$ where $$\begin{align} \alpha&=t_1+ak+ky(t_1+ak)\\ \beta &=-ky(t_0+ak) \end{align}$$

if I haven't made any mistakes.

This is easier to compute with and think about, but I don't see any way to get an explicit formula for $f_n$. I had hoped we might end up with a recursive formula for $\sum f_i$, but the coefficient of $ky$ seems to exculde any possibility of that.

This really should be a comment, not an answer, but I have no hope of typing it in a comment box.

Answer 2

Starting from Saulspatz answer: $$ f_n = \alpha+n\beta+ky\sum_{i=1}^{n-1}f_i \\\Leftrightarrow\\ \sum_{i=1}^{n}f_i=\frac{f_n -\alpha-n\beta+ky f_n }{ky} $$

Also, with $n\leftarrow n+1$:

$$ f_{n+1} = \alpha+(n+1)\beta+ky\sum_{i=1}^{n}f_i \\\Leftrightarrow\\ \frac{f_{n+1} -\alpha-(n+1)\beta }{ky}= \sum_{i=1}^{n}f_i $$

Substituting one equation into the other, we obtain $$ f_{n+1} = f_n·(k·y + 1) + β $$

This is a inhomogenous linear recurrence with constant coefficients, so you can compute the explicit form of $f_n$, and using that, probably the explicit form of $\sum f_n$.