I need to find $[x^3](1-x)^{-1}(1-2x)^6$, where $[x^3]$ means the coefficent of the $[x^3]$ term. here's what I've done:

$[x^3](1-x)^{-1}(1-2x)^6=[x^3](\sum_{k=0}^6 {6\choose k}(-2x)^k)(\sum_{m=0}^\infty {m\choose 0}x^m)$

$= \sum_{k=0}^6 {6\choose k}(-2)^k[x^{3-k} ](\sum_{m=0}^\infty {m\choose 0}x^m)$

$= \sum_{k=0}^3 {6\choose k}(-2)^k[x^{3-k} ](\sum_{m=0}^\infty {m\choose 0}x^m)$ since we need $3-k \geq 0$

$= \sum_{k=0}^3 ({6\choose k}(-2)^k {3-k\choose 0})$

$= \sum_{k=0}^3 ({6\choose k}(-2)^k)$

$= {6\choose0} + (-2){6\choose1} + (4){6\choose2} + (-8){6\choose3}$


$= -111$

But when I do the expansion on WolframAlpha, I see that $[x^0]=1$, $[x^1]=-12$, $[x^3]=-160$, so what am I doing wrong?

(I am following a similar idea to Trevor Gunn's answer in this question In how many ways the sum of 5 thrown dice is 25?)

  • $\begingroup$ Did you confuse $(-x)^k$ and $x^{-k}$? Also, did you mean $x^3$ where you wrote $x^4$? $\endgroup$ – J. W. Tanner May 22 at 0:29
  • 2
    $\begingroup$ I might have worded the question confusingly, the $[x^3]$ is not multiplication, but rather finding the coefficient of the $x^3$ term in the equation that follows $\endgroup$ – Mark Dodds May 22 at 0:36
  • 1
    $\begingroup$ Oh, then maybe you should say finding the coefficient of $x^3$ in ... $\endgroup$ – J. W. Tanner May 22 at 0:37
  • 1
    $\begingroup$ Your work is correct, and this WolframAlpha link verifies it. What specifically did you see that made you think you were wrong? $\endgroup$ – Mike Earnest May 22 at 0:55
  • 1
    $\begingroup$ @JohnOmielan looking back i think that is what has happened. Thanks for clearing that up $\endgroup$ – Mark Dodds May 22 at 0:59

As Mike Earnest confirmed in the comments, your work is correct. As I commented, and you've stated it's likely the case, the WolframAlpha results of $[x^0]=1$, $[x^1]=-12$, $[x^3]=-160$ probably come from the coefficients in the power expansion of $(1-2x)^6$ instead. You can see this directly from the first, second & fourth terms in your second last highlighted line, i.e.,



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.