# binomial thm

• Oct 8th 2008, 09:00 AM
maybeline9216
binomial thm
Write down the first three terms in the expansion, in descending powers of x, of [ax-(b/x^3)]^8. Hence evaluate the term independent of x of [5x-(1/x^3)]^8.

I dunno how to do the Hence part, making use of the terms in the expansion.
• Oct 8th 2008, 09:56 AM
ThePerfectHacker
Quote:

Originally Posted by maybeline9216
Hence evaluate the term independent of x [/B]of [5x-(1/x^3)]^8.

In this expansion we have terms of form: $\displaystyle {8\choose k} (-1)^k(5x)^k \left( \frac{1}{x^3} \right)^{8-k}$

To have a term independent of $\displaystyle x$ we require that the $\displaystyle x$ in numerator cancels out the $\displaystyle x$ in denominator. The exponent of $\displaystyle x$ in numberator is $\displaystyle k$ and the exponent of $\displaystyle x$ in denominator is $\displaystyle 3(8-k)$. We want that $\displaystyle k = 3(8-k)$ so that they cancel out. Therefore, $\displaystyle k=6$ if we solve this equation.

Therefore the indepentent term is: $\displaystyle {8\choose 6} (-1)^k 5^6$
• Oct 8th 2008, 08:04 PM
maybeline9216
Quote:

Originally Posted by ThePerfectHacker
In this expansion we have terms of form: $\displaystyle {8\choose k} (-1)^k(5x)^k \left( \frac{1}{x^3} \right)^{8-k}$

To have a term independent of $\displaystyle x$ we require that the $\displaystyle x$ in numerator cancels out the $\displaystyle x$ in denominator. The exponent of $\displaystyle x$ in numberator is $\displaystyle k$ and the exponent of $\displaystyle x$ in denominator is $\displaystyle 3(8-k)$. We want that $\displaystyle k = 3(8-k)$ so that they cancel out. Therefore, $\displaystyle k=6$ if we solve this equation.

Therefore the indepentent term is: $\displaystyle {8\choose 6} (-1)^k 5^6$

Umm....did u make use of the first part of the qn?? Cause they specify "Hence" but anyway thks