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2015/16 AHSMC Part 1 Problem 13

Problem 13 of 16

Consider the expansion (1+x+x2+⋯+x50)3=c0+c1x+c2x2+⋯+c150x150. (1 + x + x^2 + \cdots + x^{50})^3 = c_0 + c_1 x + c_2 x^2 + \cdots + c_{150} x^{150}. The value of the coefficient c50c_{50} is

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Solution

(1+x+x2+⋯+x50)3=(1+x+x2+⋯+x50)⋅(1+x+x2+⋯+x50)⋅(1+x+x2+⋯+x50). (1 + x + x^2 + \cdots + x^{50})^3 = (1 + x + x^2 + \cdots + x^{50}) \cdot (1 + x + x^2 + \cdots + x^{50}) \cdot (1 + x + x^2 + \cdots + x^{50}). The coefficient of x50x^{50} is just the number of xaxbxc=xa+b+cx^a x^b x^c = x^{a+b+c} with a+b+c=50a+b+c = 50, a,b,c∈{0,1,…,50}a,b,c \in \{0,1,\ldots,50\}. If a=0a = 0 the equation b+c=50b+c = 50 has 5151 solutions, namely (0,50)(0,50), (1,49)(1,49), …\ldots, (50,0)(50,0). Also, if a=1a = 1, the equation b+c=49b+c = 49 has 5050 solutions and so on. The number of all solutions is 51+50+⋯+1=1326. 51 + 50 + \cdots + 1 = 1326. Hence c50=1326c_{50} = 1326. The answer is (C).