Saturday, September 29, 2012


What is the coefficient of x33 in the expansion [(1+x)(1+3x3)(1+9x9)(1+27x27)]2?

 


Solution 1: Bringing the square in gives

(1+x+x+x2)(1+3x3+3x3+32x6)(1+9x9+9x9+92x18)(1+27x27+27x27+272x54)
Notice that in each factor, the power of x is 0, 1 or 2 times a power of 3. Since the base 3 representation of 33 is 10203, the only way to get x33 from this expansion is to use the term x27×x0×x6×x0. Hence, the coefficient of x33 is (2×27)×1×32×1=486.


Solution 2: Bring the square in as in Solution 1. Looking at the first three factors, the highest power we can achieve is 18+6+2=26<33, so we can't use the constant term in the fourth factor. We also can't use the x54 term in the fourth factor, thus the only possibility of achieving x33 is to use x27 in the fourth, and get x6 from the first three. Now consider the third factor, we can't use x9 or x18, so we have to use the constant term. From the first 2 factors, it is clear that we can only use 1from the first factor and x6 from the second factor. Hence, the coefficient of x33 is: 1×32×1×(2×27)=486.

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