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In the polynomial $$ (x-1)(x^2-2)(x^3-3) \ldots (x^{11}-11) $$ what is the coefficient of $x^{60}$?

I've been trying to solve this question since a long time but I couldn't. I don't know whether opening the brackets would help because that is really a mess. I have run out of ideas. Would someone please help me to solve this question?

AryanSonwatikar
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shsh23
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  • Think of how many ways the terms can form a term $x^{60}$ ... It could be $x^1 \cdot x^{59}$, or $x^2 \cdot x^{58}$, or even $x^3 x^8 x^{49}$ ... Can you think of a systematic way to create $x^{60}$ like this? – Matti P. Nov 05 '19 at 10:38
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    Old question, see here. Also on this site there are several such question. Please have a look, say here. Next time, please search before posting. – Dietrich Burde Nov 05 '19 at 10:39
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    @DietrichBurde it's not easy to search for mathematical expressions (unless you know to use approach 0 or something like it), and there isn't a lot of text to go off of for these problems. I don't think that there's much reason to assume that the asker didn't at least try searching. – Ben Grossmann Nov 05 '19 at 10:44
  • @Omnomnomnom I agree, that it is not always immediate to see these duplicates, but honestly my experience on this site is that several users are not even aware that it is possible to find the same question with several beautiful answers. This is true in particular for these "standard questions". One of the goals for students also is to learn how to search. – Dietrich Burde Nov 05 '19 at 10:47
  • @Aryan The previous edit was better actually. – shsh23 Nov 05 '19 at 10:48
  • @Shashwat1337 I only edited the tags. In case you are unaware, such a problem can be solved easily with the knowledge of Binomial Theorem. Edit: When I proceeded to edit the tags, Matti P. hadn't edited it. So my edit applied to your initial version. – AryanSonwatikar Nov 05 '19 at 10:55
  • @Aryan Sorry my bad. Please don't mind. – shsh23 Nov 05 '19 at 10:59
  • @Shashwat1337 It's all right. – AryanSonwatikar Nov 05 '19 at 11:01

3 Answers3

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The highest exponent possible is $1+2+ \cdots + 11 = 66$.

Now, to create the exponent $60$, you can only leave out the factors containing $(1,2,3),(2,4),(1,5)$ and $6$. Hence,

  • $1+2+3 \Rightarrow$ gives coefficient $(-1)(-2)(-3) = -6$
  • $2+4 \Rightarrow$ gives coefficient $(-2)(-4) = 8$
  • $1+5 \Rightarrow$ gives coefficient $5$
  • $6 \Rightarrow$ gives coefficient $-6$

Summing up gives $13-12 = \boxed{1}$.

trancelocation
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2

Hint : $1+2+3 +...+11= \frac {11×12}{2} =66 $ so we must find how we can construct number $6=6+0=5+1=4+2=3+3=1+2+3$ and note that $3+3$ impossible.

amir bahadory
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1

Expanding the brackets is definitely not the way to go, but thinking about what would happen if you did is helpful. Every term in the expanded polynomial will come from multiplying one term from each bracket (e.g. the highest degree term will come from multiplying $\,x \cdot x^2 \cdot x^3 \cdot x^4 \dots \cdot x^{11}\,$), and since each bracket is a binomial that doesn't leave many options.

The first thing to notice is that the aforementioned highest-degree term will be $\,x^{66};\,$ thus, to get a term with $\,x^{60},\,$ we will need to use MOST BUT NOT ALL of the $\,x$'s in the factors. We need to reduce the maximum degree by $\,6\,$; one obvious way to do that is to not use the $\,x^6\,$ in the sixth factor, so instead we use the $\,-6\,$ from that bracket. However, there are other ways, e.g. not using the $\,x^2\,$ or the $\,x^4\,$ from the second and fourth factors, respectively. Hopefully by now you're getting the sense that what we need to do is come up with all the different ways to make $\,x^6\,$ with any of the powers available to us, and then for each way find the corresponding coefficient.

There are not that many ways of making $\,x^6;\,$ note that the most we can use is three powers of $\,x\,$, as the minimum power would then be $\,x\cdot x^2\cdot x^3=x^6.\,$ The list of ways is: $\,x^6, x\cdot x^5, x^2 \cdot x^4,\,$ and $\,x\cdot x^2\cdot x^3.\,$

Now we just need to find the coefficient of each term that EXCLUDES one of the above combinations. As an example, let's consider the term that excludes $\,x^2\,$ and $\,x^4.\,$ Written out in detail, this term would be

$$(x)(-2)(x^3)(-4)(x^5)(x^6)(x^7)(x^8)(x^9)(x^{10})(x^{11}) = 8x^{60}$$

Similarly, it should be easy to see that the coefficients for the terms excluding the other combinations are:

  • For $\,x^6,\,$, the coefficient will just be $\,-6\,$;
  • For $\,x\cdot x^5,\,$ the coefficient will be $\,(-1)(-5)=5\,$;
  • For $\,\,x\cdot x^2\cdot x^3,\,$ the coefficient will be $\,(-1)(-2)(-3)=-6.\,$

Thus the four terms that will have $\,x^{60}\,$ are $\,-6x^{60}, 5x^{60}, 8x^{60}\,$ and $\,-6x^{60}.\,$ The sum of these will be just $\,x^{60},\,$ so the answer to the question is $\,\boxed{1}.\,$

A.J.
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