1. ## without expanding

without expanding, determine the ff

term free of y in (2x^2 y^4 + x^3 y^-2) ^15

tnx

2. hi kabayan!

find $n$ such that $y^{4(15-n)}y^{-2n}=y^0$

so basically, you only need to solve for $4(15-n)+(-2n)=0$

once you've solve this, then find $\left(\begin{array}{r}15 \\ n \end{array}\right)(2x^2)^{15-n}(x^{3n})$

3. Hello kalagota,

Originally Posted by kalagota

once you've solve this, then find $\left(\begin{array}{r}15 \\ n \end{array}\right)(2x^2)^{15-n}(x^{3n})$
Instead of using an array, you can do ${15 \choose n}$ $${15 \choose n}$$ (there is another way but I don't remember)

4. Originally Posted by Moo
(there is another way but I don't remember)
$$\binom{15}{n}$$ gives $\binom{15}{n}$

5. Originally Posted by Isomorphism
$$\binom{15}{n}$$ gives $\binom{15}{n}$
this doesn't seem to work on my editor/compiler..