Thread: Binomial Series

Use the binomial series to get a 3 term series for (2+4rootx)^3 and find the valid values for x.

What do u do with the 2?

Ok. You gotta change it to the form $\displaystyle (1+x)^n$ so...

$\displaystyle (2 + 4 \sqrt{x})^2 = (2(1 + 2\sqrt{x}))^3 = 2^3 (1+2\sqrt{x})^3$

Then plug this into the binomial series which is...

$\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

With $\displaystyle n=3$ and $\displaystyle x=2\sqrt{x}$.

The binomial series is valid when $\displaystyle |x|<1$ so find $\displaystyle x$ such that $\displaystyle |2\sqrt{x}|<1$

Thanks ALOT, sorry for late reply...