# Find the coefficient of the power series $[x^3](1-x)^{-1}(1-2x)^6$

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 $$3-k \geq 0$$

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

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

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

$$=1-12+60-160$$

$$= -111$$

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?)

• Did you confuse $(-x)^k$ and $x^{-k}$? Also, did you mean $x^3$ where you wrote $x^4$? – J. W. Tanner May 22 at 0:29
• 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 – Mark Dodds May 22 at 0:36
• Oh, then maybe you should say finding the coefficient of $x^3$ in ... – J. W. Tanner May 22 at 0:37
• Your work is correct, and this WolframAlpha link verifies it. What specifically did you see that made you think you were wrong? – Mike Earnest May 22 at 0:55
• @JohnOmielan looking back i think that is what has happened. Thanks for clearing that up – 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.,

$$=1-12+60-160$$