Question

find the term that contains x^8 y^9 in the expansion of (x^2+3y^3)^7

Answer 1

first term in the expansion:\[\binom{7}{0}\left(x^{2}\right)^7\left(3y^3\right)^0 = x^{14}\]
second:\[\binom{7}{1}\left(x^2\right)^6 \left(3y^3\right)^1 = 21x^{14}y^3\]Seems that the one which brings the 9th power is where \(3y^3\) is raised to the third. So the fourth term. Find out what that is.

Answer 2

@ParthKohli

Answer 3

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