"Solving Coefficient of x^n in Binomial Expansion

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To find the coefficient of x^n in the binomial expansion of (1-x)^(-6), the generalized binomial series is used, expressed as (1 - x)^m = Σ (m choose k) x^k. The coefficient for x^n is determined by setting -6 - k = n, leading to k = -6 - n. The formula for the coefficient simplifies to -6(-7)...(-11-n)/(-6-n)!. The discussion also touches on the differences in binomial series formulas for positive versus negative exponents and confirms that the order of terms in the expansion does not affect the result due to the commutative property of addition. Further hints are requested for finding coefficients of x^6 and x^7 in a different polynomial expansion.
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Homework Statement



Find, in the simplest form, the coefficient of x^n in the binomial expansion of (1-x)^(-6).

Homework Equations





The Attempt at a Solution



i am not sure how to go about with this.
 
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are u sure you have to find a cofficient contain x^n?

because u should have a specific value for n, so that you could find the cofficient infront of it, or you should at least have which term you are looking for.
 
Willian93 said:
are u sure you have to find a cofficient contain x^n?

because u should have a specific value for n, so that you could find the cofficient infront of it, or you should at least have which term you are looking for.

Yes, that's the question. Maybe it's asking for the coefficient of for any term in the expression.
 
Since the exponent, -6, is not a positive integer, you will need to use the generalized binomial series:
(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}

Here, of course, a= 1, b= -x, and m= -6 so this is

(1- x)^{-6}= \sum_{k=0}^\infty \frac{-6(-7)\cdot\cdot\cdot(-5-k)}{k!}(-1)^kx^{-6-k}

You, apparently, are asked for the coefficient when -6-k= n or when k= -6-n.
 
HallsofIvy said:
Since the exponent, -6, is not a positive integer, you will need to use the generalized binomial series:
(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}

Here, of course, a= 1, b= -x, and m= -6 so this is

(1- x)^{-6}= \sum_{k=0}^\infty \frac{-6(-7)\cdot\cdot\cdot(-5-k)}{k!}(-1)^kx^{-6-k}

You, apparently, are asked for the coefficient when -6-k= n or when k= -6-n.

thanks

Is this answer the most simplified?

\frac{-6(-7)...(-11-n)}{(-6-n)!}

the general formula for binomial series for (a+b)^n is different when n is a positive integer and when n is a fractional or negative value?

(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}

Does it matter if the powers(k and n-k) for a and b is swapped since a+b is commutative?

This is the continuation of the question:

Hence, find the coefficient of x^6 and x^7 in (1+2x+3x^2+4x^3+5x^4+6x^5+7x^6)^3
 
Last edited:
any further hints on this question?
 
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