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I have following equations:

$$\frac{1}{((a_1 - 1) f + 1)} + \frac{1}{((a_2 - 1) f + 1)} -k = 0$$ $$\frac{1}{((a_1 - 1) f + 1)} + \frac{1}{((a_1 - 1) f + 1)} + ... + \frac{1}{((a_n - 1) f + 1)} -k = 0$$

which can be written as:

$$(\sum_{i=1}^n \frac{1}{((a_i - 1) f + 1)}) - k = 0; n>=2$$

For each of $n$ values, I get polynomials of 2nd, 3rd, ... nth grade.

Is it easy to generalize this formula to get it somehow in this form?:

$$(\sum_{i=0}^n f^{i}(unknown))=0; n>=2$$

n=2 (and assuming $b_n=a_n-1$) I get:

$$b_1 b_2 f^2 (-k)+f \left(b_1 (-k)-b_2 k+b_1+b_2\right)-k+2=0$$

n=3 (and assuming $b_n=a_n-1$) I get:

$$b_1 b_2 b_3 f^3 (-k)+f^2 \left(b_1 b_2 (-k)-b_3 b_2 k-b_1 b_3 k+b_1 b_2+b_3 b_2+b_1 b_3\right)+f \left(b_1 (-k)-b_2 k-b_3 k+2 b_1+2 b_2+2 b_3\right)-k+3=0$$

n=4 (and assuming $b_n=a_n-1$) I get:

$$b_1 b_2 b_3 b_4 f^4 (-k)+f^3 \left(b_1 b_2 b_3 (-k)-b_1 b_4 b_3 k-b_2 b_4 b_3 k-b_1 b_2 b_4 k+b_1 b_2 b_3+b_1 b_4 b_3+b_2 b_4 b_3+b_1 b_2 b_4\right)+f^2 \left(b_1 b_2 (-k)-b_3 b_2 k-b_4 b_2 k-b_1 b_3 k-b_1 b_4 k-b_3 b_4 k+2 b_1 b_2+2 b_3 b_2+2 b_4 b_2+2 b_1 b_3+2 b_1 b_4+2 b_3 b_4\right)+f \left(b_1 (-k)-b_2 k-b_3 k-b_4 k+3 b_1+3 b_2+3 b_3+3 b_4\right)-k+4=0$$

And so on.. Not sure if there is some general solution to this, so I can easily construct equation without the need to multiply and group params.

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  • $\begingroup$ First of all, your equations would look much simpler if you make a substitution $b_n=a_n-1$. $\endgroup$
    – yarchik
    Jul 22, 2019 at 12:09
  • $\begingroup$ @yarchik thanks for suggestion. will try now :) $\endgroup$ Jul 22, 2019 at 12:15

1 Answer 1

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OK, after simplifying by using suggestion from comments, I think I have it now:

$$\sum_{i=0}^nf^i\binom{\{b_1,...,b_n\}}{i}(-k+n-i)$$

Just don't know if this is proper way to express all $i$ combinations of $b_1...b_n$ elements

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