MATH QUESTION.... Find the coefficient of x^39 in (x-1)^41?

Use binomial expansion,

(x+y)^n = nC0 (x^n)(y^0) + nC1 (x^n-1)(y^1) + nC2 (x^n-2)(y^2) + ...... +nCn (x^0)(y^n)

here, y = -1

and, n = 41

so, 3rd term in expansion = 41C2 (x^39)((-1)^2)

= 41! / (39! 2!) x^39

= 820 x^39

coefficient = 820

[nCr = n! / r! (n-r)!

and, n! = n(n-1)(n-2)...(3)(2)(1)]

Since you have an essentially correct answer to your previous question, I can only assume that the difficulty that remains is in the numerical interpretation of the binomial coefficient 41C2.

In general terms, if you have a binomial expansion (a - b)^n, the result is a series of (n + 1) terms of the general form

(a - b)^n = ∑ [nCk].[a^(n-k)].[(-b)^k] where k = 0 to n

The definition of the binomial coefficient is nCk = n!/[k! (n - k)!], where the exclamation signs indicate a factorial k! = 1 * 2 * 3 * ... * (k - 2) * (k - 1) * k. When you come to calculate nCk in a specific case, for example n = 41, k = 2 and n- k = 39, it produces

41C2 = 1*2*3*4*5* ... *38*39*40*41

. . . . . --------------------------------------------

. . . . . [1*2]*[1*2*3*4*5* ... *37*38*39]

Since the factorials all start from 1, the early terms in the longer expansions cancel out to leave

41C2 = 40*41/[1*2] = 20*41 = 820

The binomial terms a and b may themselves contain numerical coefficients, which will in general combine to make the coefficient of x more complicated, but in this case the coefficients of (x - 1) are 1 and -1, so their only effect is to change the sign of alternate terms.

Make it a simpler problem. Consider (x-1)^3.

(x-1)(x-1)(x-1)

Now consider what needs to be multiplied when x^3:

(x )(x )(x )

There is only one possible way and the coefficient is 1.

Similarly the coefficient of x^0, the constant, is -1.

Now consider x^2:

(x )(x )( -1)

(x )( -1)(x )

( -1)(x )(x )

There are three ways. Each equal -1x^2. Therefore the coefficient is -3.

Similarly the coefficient of x is 3.

So each possible way you can multiply to get x^39 has x^39*(-1)^2 as the value (the exponents add to 41) which is x^39.

There are 39 choose 41 possible ways to multiply out to get x^39. That is equivalent to 2 choose 41.

Therefore the outcome is that the coefficient of x^39 is 2C41.

You need to understand how the binomial theorem works.

You can then find all the coefficients such as x^40, x^39, x^38, etc down to x^2, x.