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gavmur [86]
2 years ago
5

Please help and no links please

Mathematics
1 answer:
gladu [14]2 years ago
7 0

Answer:

sorry i can't clear

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lys-0071 [83]

\displaystyle&#10;(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

----------------------------------------------------

\displaystyle&#10;(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\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

6 0
3 years ago
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