find the term containing 4^10 in the expression of (3x+y)^12
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In the binomial expansion there will be two terms containing a tenth power. The coefficients in the expansion of (a+b)^12 are 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1. The terms with coefficient 66 are a^10b^2 and a^2b^10. By analogy then, where a is replaced by 3x and b by y, we have 66*3^10x^10y^2=3897234x^10y^2 and 66*3^2x^2y^10=594x^2y^10. So there can be no 4^10 (1048576) term.

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