(2x^2 - 3/x)^18

how can I find the 16th term and follow up by finding the term that is independent of x?

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- May 15th 2006, 02:57 PMStuartterms and others
(2x^2 - 3/x)^18

how can I find the 16th term and follow up by finding the term that is independent of x? - May 15th 2006, 03:59 PMThePerfectHackerQuote:

Originally Posted by**Stuart**

$\displaystyle {18 \choose 15}a^{15}b^3$

Over here $\displaystyle a=2x^2,b=-3/x$

Thus,

$\displaystyle {18 \choose 15}(2x^2)^{15}(-3/x)^3=-816(2^{15}x^{30})\cdot \frac{3^3}{x^3}=-815\cdot 2^{15} \cdot 3^3\cdot x^{27}$ - May 15th 2006, 04:05 PMStuart
Ok well I am sure that K = 12 and N = 18

so to set it up is 18c12 * (2x^2)^18-12 * (-3/x)^12

Right and follow through with that you gave for the 16th term? If not what does the 18 over 15 mean? - May 15th 2006, 04:26 PMThePerfectHackerQuote:

Originally Posted by**Stuart**