Does it Have it?

Let the expression be $P (x)$ . We get the highest power of $x$ in the expansion of $P(x)$ by multiplying all the $x^k$ together, where $k=1,2,3...20$ . That is $\displaystyle \sum_{k=1}^{20} k = \frac {20(21)}2 = 210$ and the coefficient of $x^{210}$ is $1$ . Similarly, we get the term with $x^{209}$ by multiplying the $-1$ of $x-1$ with the rest of $x^k$ , that is $-1\cdot x^{209}$ giving the coefficient $a_{209}=-1$ . Similarly, the coefficient of $x^{208}$ is multiplying $-2$ of $x^2-2$ with the rest of $x^k$ , that is $a_{208} = -2$ . For the coefficient of $x^{203}$ , we have:

$a_{203} = -7 + (-1)(-6)+(-2)(-5)+(-3)(-4) + (-1)(-2)(-4) = -7 + 6 + 10 + 12 - 8 = \boxed{13} \quad \small \color{#3D99F6} \text{See note.}$

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Note:

$\begin{aligned} P(x) & = \prod_{k=1}^{20} (x-k) = (x^7 - 7) \prod_{k=1, k \ne 7}^{20} (x-k) = (x^7 - 7)\bigg(x^{203} - 203x^{202} + \cdots - \prod_{k=1,k \ne 7}^{20} k \bigg) = - 7 x^{203} + \cdots \end{aligned}$

Similarly,

$P(x) = (x-1)(x^6 - 6)(x^{203} + \cdots) = (x-1)(-6x^{203} + \cdots) = 6 x^{203} + \cdots \\ P(x) = (x^2-2)(x^5-5)(x^{203} + \cdots) = 10x^{203} + \cdots \\ P(x) = (x^3-3)(x^4-4)(x^{203} + \cdots) = 12x^{203} + \cdots \\ P(x) = (x-1)(x^-2)(x^4-4)(x^{203} + \cdots) = -8x^{203} + \cdots$

Therefore, $a_{203} = -7+6+10+12-8 = \boxed{13}$ . Note that the sum of the product number is $-7, - 1-6, -2-5, -3-4, -1-2-4$ .

The coefficient of the $x^{203}$ term can be found by finding the sum of the coefficients of $x^{203}$ from all the possible ways $203$ can be made by all different integers from $1$ to $20$ . Since the highest number that can be made with these integers is $\frac{20(20 + 1)}{2} = 210$ , another approach is to use up all the different integers from $1$ to $20$ except for the different integer combinations that can add up to $210 - 203 = 7$ , which are $7$ , $1$ and $6$ , $2$ and $5$ , $3$ and $4$ , and $1$ and $2$ and $4$ . The possibilities are then:

$(x^7 - 7) \cdot [(x - 1) \dots (x^6 - 6)(x^8 - 8) \dots (x^{20} - 20)] = (x^7 - 7)(x^{203} + \dots) = x^{210} + \dots - 7x^{203} + \dots$

$(x - 1)(x^6 - 6) \cdot [(x^2 - 2) \dots (x^5 - 5)(x^7 - 7) \dots (x^{20} - 20)] = (x^7 + \dots + 6)(x^{203} + \dots) = x^{210} + \dots + 6x^{203} + \dots$

$(x^2 - 2)(x^5 - 5) \cdot [(x - 1)(x^3 - 3) \dots (x^4 - 4)(x^6 - 6) \dots (x^{20} - 20)] = (x^7 + \dots + 10)(x^{203} + \dots) = x^{210} + \dots + 10x^{203} + \dots$

$(x^3 - 3)(x^4 - 4) \cdot [(x - 1)(x^2 - 2) \dots (x^5 - 5) \dots (x^{20} - 20)] = (x^7 + \dots + 12)(x^{203} + \dots) = x^{210} + \dots + 12x^{203} + \dots$

$(x - 1)(x^2 - 2)(x^4 - 4) \cdot [(x^3 - 3)(x^5 - 5) \dots (x^{20} - 20)] = (x^7 + \dots - 8)(x^{203} + \dots) = x^{210} + \dots - 8x^{203} + \dots$

The sum of the possible $x^{203}$ term coefficients is $-7 + 6 + 10 + 12 - 8 = \boxed{13}$ .

Divide the whole thing by $x^{210}.$ The coefficient of $x^{203}$ in the original product is the coefficient of $x^{-7}$ in $\left( 1-\frac1{x} \right) \left( 1 - \frac2{x^2} \right) \cdots \left( 1-\frac{20}{x^{20}} \right).$ Let $y = \frac1{x},$ so we want the coefficient of $y^7$ in $(1-y)(1-2y^2)\cdots(1-20y^{20}).$ Getting $y^7$ in the product involves a sum of distinct positive integer exponents equal to $7.$ There are five ways to do this: $7, 1+6, 2+5, 3+4, 1+2+4.$ These yield the coefficients $-7, (-1)(-6), (-2)(-5), (-3)(-4), (-1)(-2)(-4)$ respectively. These add up to $\fbox{13}.$

Expanding an expression like the one given is done by combining all possible ways to select the first or the second term in each factor.

The highest possible power is $x^{210}$ which you get by multiplying all the $x$ terms. To get $x^{203}$ you need to skip seven $x$ , which can be done by skipping $x^7$ , or by skipping $x^6$ and $x$ , or any other combination of 1 to 7 that sums to 7.

There are only 5 ways to combine the integers 1 to 7 and get 7:

• $7$
• $6 + 1$
• $5 + 2$
• $4 + 3$
• $4 + 2 + 1$

For each of these you get one term of $k \times x^{203}$ . The value of $k$ depends on the alternative (non $x$ ) terms in the given expression. For skipping $x^{7}$ we get $-7$ , for skipping $x^{4}$ and $x^{3}$ , we get the combination of $-4$ and $-3$ . Calculating them one by one we get:

• $-7$
• $-6 \times -1 = 6$
• $-5 \times -2 = 10$
• $-4 \times -3 = 12$
• $-4 \times -2 \times -1 = -8$

Summing these we get the answer:

$-7 + 6 + 10 + 12 + -8 = \boxed{13}$

Using $Mathematica$

Coefficient[Product[x^n-n,{n,20}],x^203]

returns $13$

The product of these binomials could be imagined as a game of switching " power up " or " constant up " where in power up you will add the location of the switch to the exponent of $x$ and in constant up , you will multiply the constant to the power of $x$ , for example:

First switch $x-1$ :

• power up : add one to the exponent of $x$

• constant up : add multiply the constant $-1$ to the power $x$

Second switch $x-1$ :

• power up : add two to the exponent of $x$

• constant up : add multiply the constant $-2$ to the power $x$

Suppose we are dealing with 20 series of switches where power up s will add to the exponent of $x$ (depending on the location of the switch as exampled) and constant up will be multiplied to the resulting power of $x$ .

Now, if power ups are maximized, it will result to 210 as the exponent of $x$ . However, we only need 203, thus 7 exponents shall be removed. Hence, instead of powering up, we need to constant up some of the selected switches totaling to 7 exponents. The plausible candidates are:

A. 7th switch $->$ constant up: $-7$

B. 6th and 1st switch $->$ constant up: $(-6)(-1) = 6$

C. 1st, 2nd and 4th switch $->$ constant up: $(-1)(-2)(-4) = -8$

D. 2nd and 5th switch $->$ constant up: $(-2)(-5) = 10$

E. 3rd and 4th switch $->$ constant up: $(-3)(-4) = 12$

Since these are mutually exclusive events, hence, we shall add them, thus, $-7 + 6 + (-8) + 10 + 12 = \boxed{13}$

In short, one of the terms in the expansion is $13x^{203}$ .