Binomial Series

**TreePerson** · March 28th 2010, 01:34 PM

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?

**Deadstar** · March 28th 2010, 04:03 PM

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

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

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

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

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

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

**TreePerson** · April 16th 2010, 09:32 PM

Thanks ALOT, sorry for late reply...

Math Help - Binomial Series

LinkBack

Thread Tools

Display

Binomial Series

Similar Math Help Forum Discussions

binomial Series Sum.

Binomial series

Binomial series?

Binomial Series to find a Maclaurin Series

Binomial series

Search Tags