what is the coefficient of x^49 in the expansion of (x-1)(x-2)(x-3)............(x-50)

Question

The choices are.

(a) -2250

(b) -1275

(c) 1275

(d) 2250

(e) -49

Answer 1

(x-1)(x-2)(x-3)............(x-50) = x⁵⁰ + a₁x⁴⁹ + a₂x⁴⁸ + ...... + a₄₈x² + a₄₉x + a₅₀

 

For (x-1)(x-2)(x-3)............(x-50) = 0, a1 is minus the sum the roots of the equation.

Since the roots are 1, 2, 3, ..., 50

a1 = -(1 + 2 + 3 + ...+ 49 + 50)

    = -50*51/2 = -1275

Answer 2

I am not sure how to do this with the Binomial Theorem, but I started to do the multiplcations.

 

First I did it in pairs, so (x-1)(x-2)  and multiplied (x-3)(x-4)....   

 

If n = the degree (the highest power), then the coefficient of the (n-1)th term (here the x¹) is just the sum of the factors.  Meaning:  (x-1)(x-2) = x² - 3x + 2       Notice, the (n-1)th term is just the (-1) of the first factor + (-2) of the second factor.

 

Do it again with (x-3)(x-4) = x²-7x+12    Notice, the (n-1)th term has a coefficient of (-3) + (-4).

 

In the next layer, take these quadratics in pairs, and get

 

(x² - 3x + 2)(x²-7x+12) = x⁴ - 10x³ +35x² - 50x + 24

 

Here, the (n-1)th coefficient is -10.  Notice that this = (-1) + (-2) + (-3) + (-4)  <-- the sum of the factors involved)

 

 

I did one more layer, this time using the first 8 factors, to find that  

 

(x⁴ - 10x³ + 35x² - 50x +24)(x⁴ - 26x³ + 485x² -18930x+168) =

 

x⁸ - 36x⁷ + ....  

 

(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+(-7)+(-8)=-36

 

Again, notice that the (n-1)th coefficient is the sum of all of the first n factors (8 factors in this case)

 

Notice that the highest power when we get finished multiplying all of these will be n = 50, so the (n-1)th term will be the 49th term.  This is the coefficient you are looking for.

 

So, if you add (-1)+(-2)+(-3)+(-4)+ ... +(-47)+(-48)+(-49)+(-50) = -1275

 

Answer is (B)