<p>Sorry I didn't know what other place would fit this thread. It is a sample practice problem for a high school math competition coming up tomorrow. </p>
<p>"Find the constant term in the expansion of (3x^2 + 1/x)^6"</p>
<p>If anyone could help me it'd be awesome!!!</p>
<p>This is easy (X^2)^n =(1/x)^(6-n) so x=2. So answer = 6 choose 2 times 3^2= 15 times 9 = 135.</p>
<p>From the binomial theorem, we know that each term in the binomial expansion is equal to some constant multiplied by some power of (1/x), multiplied by some power of (3x^2). Obviously, a term in the binomial expansion will be constant when (1/x) is being raised to a power twice as big as the power on (3x^2), because the x’s will cancel.
The only time this happens is in the (3x^2)^2 * (1/x)^4 term. The binomial coefficient on that term is (6 choose 4) = 15, but we need to multiply that by 9 because (a^2) = (3x^2)^2 = 9x^4. The x^4 will cancel, but the 9 won’t, so the constant term is 9*15 = 135. </p>
<p>Thanks a lot guys :)</p>
<p>I also have one more question. </p>
<p>“How many non-negative integer solutions are there to the equation x + y + z + w = 10” ?</p>
<p>@blackhole22 </p>
<p>I swear people don’t even know what Google is anymore</p>
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<p>@blackhole22 You should be on the AoPS forums for contest math questions… constantly being involved on AoPS will help your skills!</p>