Respuesta :

[tex] \displaystyle (x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k [/tex]

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[tex] \displaystyle (x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\ n=5\\ k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10 [/tex]

It's 10.

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[tex] \displaystyle (3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\ n=5\\ k=4\\\\ \binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90[/tex]

It's 90

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