# terms and others

• May 15th 2006, 02:57 PM
Stuart
terms 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 PM
ThePerfectHacker
Quote:

Originally Posted by Stuart
(2x^2 - 3/x)^18

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

The 16th term of $\displaystyle (a+b)^{18}$ is,
$\displaystyle {18 \choose 15}a^{15}b^3$
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 PM
Stuart
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 PM
ThePerfectHacker
Quote:

Originally Posted by Stuart
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?

It is another way of writing 18C15