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Ceedero1015 · Dec 12, 2008, 09:43 AM

Find the coefficients of X^12y^6 in the expansion of (x^3-3y)^10

Find the constant term in the expansion of (y-1/2y)^10

galactus · Dec 13, 2008, 03:53 PM

$(x^{3}-3y)^{10}$

Here is one way to go about it by listing out the terms.

The first term will be $x^{30}$

The coefficient of the second term will be the exponent from the first term.

$30x^{27}y$

Reduce the exponents of x by 3 and add 1 to the exponent of y.

Now, To find the next coefficient, multiply the coefficient by the exponent and divide by 1 more than the exponent of y. 30*27/2=405

So, the next term is $405x^{24}y^{2}$

The next would be 405*24/3=3240

$3240x^{21}y^{3}$

Continue in this manner until you get to y^6. Not too much further.

So far, we have $x^{30}-30x^{27}y+405x^{24}y^{2}-3240x^{21}y^{3}+....$

Remember to alternate your signs.

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For the second one, is that $(y-\frac{1}{2}y)^{10}$ ?

If so, look at it. It is just $(\frac{y}{2})^{10}$ . Now, can you see the constant term?