Consider the product series:

$(x-a)$ $(x^{2} -b)$ .......$(x^{14} -n)$

I want to express the result in this form:

$x^{1+2+3+.....14}$ + (some constant)$x^{1+2+3+...14-1}$ +.......so on

My question is how to find the general expression for the coefficients for every powers of x(which are decreasing)?

I have no idea please help me

Is there any formula which can help me?

  • $\begingroup$ I would start with setting $a, b, \ldots,n$ to $1$ first and see what you get. Then set them all to $-1$, and see what you get. That's difficult enough, and at the same time ought to give you a good idea of how to attack the general problem. $\endgroup$ – Arthur Jan 31 '18 at 7:33
  • $\begingroup$ @Arthur Yes I completely understand but I need to express it in decreasing powers of x and all I want is some formula or generalization for it's respective coefficients $\endgroup$ – David Jan 31 '18 at 7:36
  • $\begingroup$ @Arthur I think setting them to 1 is an assumption sir. Can't we do it without any assumption? $\endgroup$ – David Jan 31 '18 at 7:41
  • $\begingroup$ You are entirely correct that it is an assumption. As I said, I thiunk you should do it with some restrictive assumptions first, just to see what happens. Then once you've done that, you can try to do the whole thing. $\endgroup$ – Arthur Jan 31 '18 at 7:46
  • $\begingroup$ @Arthur But sir in that case the whole question will become different. But I want to thank you for giving me the idea...Hope someone solves it $\endgroup$ – David Jan 31 '18 at 7:51

No, there isn't a general formula. You know that the leading term is $x^105$ because $\sum_{k=1}^{14}{k}=105.$ Now the next term must be $-ax^{104}$ because there's only one way to get $x^{104}.$ Similarly, the third term is $-bx^{103}.$ But when we get to $x^{102}$ things change. We might have multiplied all the variables but $x^3$ or all the variables but $x$ and $x^2$, so the coefficient is $ab-c$. Things will get more complicated as you go along. There are $2^{14}=16,384$ ways to pick which constants to multiply together, and only 106 terms among which to distribute the products, so it's going to get messy.

May I ask why you want to know the individual terms? It seems like you've got a pretty simple formula now.

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  • $\begingroup$ Sir, I want to know it because I require it for a problem in my exercise $\endgroup$ – David Jan 31 '18 at 7:48
  • $\begingroup$ I still believe there can be some way sir $\endgroup$ – David Jan 31 '18 at 7:54
  • $\begingroup$ David, there has to be something more to this problem. This would be far too arduous as an exercise. No teacher would make you compute 16,000 terms -- he wouldn't be able to grade it! What exactly is the problem statement? $\endgroup$ – saulspatz Jan 31 '18 at 7:55
  • $\begingroup$ It was given as a bounty problem. I guess we need to invent some formula for it if it is possible. The problem was the same written here. I just want to devise some formula not because of the reward but because of the love for mathematics $\endgroup$ – David Jan 31 '18 at 8:06
  • $\begingroup$ I wish you good luck, but I'm not optimistic. $\endgroup$ – saulspatz Jan 31 '18 at 19:17

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