Expand $(3x^2+y)^5$ Using binomial theorem we have $\sum_{k=0}^{n} x^ky^{n-k}$ where n =5.
I tried setting it up like this:
${5 \choose 0}3x^{2k}y^{n-k}+.......+{5 \choose 5}3x^{2k}y^{n-k}$
for ex. k=2:
${5 \choose 2} 3x^4y^3 = 10 \times 3x^4y^3 = 30x^4y^3$
My answer was way off. My powers were all correct but my coefficients were way off, not even in the same ballpark. How do I go about calculating this?
 A: This might prove to be very useful to you. Although you have been correctly guided in the comments section. You need to use parentheses correctly.
A: We have: ${5\choose{0}}(3x^{2})^{5}+{5\choose{1}}(3x^{2})^{4}(y)+{5\choose{2}}(3x^{2})^{3}(y)^{2}+{5\choose{3}}(3x^{2})^{2}(y)^{3}+{5\choose{4}}(3x^{2})(y)^{4}+{5\choose{5}}(y)^{5}$
$=(1)(243x^{10})+(5)(81x^{8})(y)+(10)(27x^{6})(y^{2})+(10)(9x^{4})(y^{3})+(5)(3x^{2})(y^{4})+(1)(y^{5})$
$=243x^{10}+405x^{8}y+270x^{6}y^{2}+90x^{4}y^{3}+15x^{2}y^{4}+y^{5}$
A: For the coefficients I would simply use Pascal's triangle. Then you just have to include the powers of $(3x^2)^m$ and $(3x^2)^n$ with $n+m = 5$
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For a fast calculation of the triangle, observe that the left and right most numbers are always $1$ and those in between equal the sum of the two digits above it. 
A: You have to raise powers of $3x^2$
