# Binominal series

• Jun 24th 2006, 08:08 AM
Paps
Binominal series
hi im new here and not very good with math terms of english words :)

i have this to make and i just dont know how i tried but it just wouldnt happen

Write (find) the first 3 terms in binomianl series to calculate approximate value for n-th root ; n=3.

28^(1/n) ?

• Jun 24th 2006, 09:29 AM
Soroban
Hello, Paps!

Welcome aboard!

Quote:

Find the first 3 terms in binomianl series to calculate approximate value of: $28^{\frac{1}{3}}$

The binomial series looks like this:

$(a + b)^n\;=\;a^n + n\!\cdot\!a^{n-1}b + \frac{n(n-1)}{2!}a^{n-2}b^2 \,+$ $\frac{n(n-1)(n-2)}{3!}a^{n-3}b^3 + \hdots + b^n
$

We have: . $28^{\frac{1}{3}}\;=\;(27 + 1)^{\frac{1}{3}}$ . . . where: $a = 27,\;\;b = 1,\;\;n = \frac{1}{3}$

The first three terms are: . $(27 + 1)^{\frac{1}{3}}\;=\;27^{\frac{1}{3}} + \frac{1}{3}(27^{-\frac{2}{3}})(1) + \frac{\frac{1}{3}(\frac{1}{3} - 1)}{2!}(27^{-\frac{5}{3}})(1^2)$

. . $= \;\sqrt[3]{27} + \frac{1}{3(\sqrt[3]{27})^2} + \frac{\frac{1}{3}(-\frac{2}{3})}{2(\sqrt[3]{27})^5} \;=$ $3 + \frac{1}{3\cdot9} + \frac{-\frac{2}{9}}{2\cdot243} \;= \;3 + \frac{1}{27} - \frac{1}{2187}$

Therefore: . $\boxed{\sqrt[3]{28} \;\approx \;3.03657979}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is a very good approximation!

The actual value is: . $\sqrt[3]{28}\;=\;3.036588972$

Our answer gives us: . $3.03657979^3\;=\;27.9997...$

• Jun 24th 2006, 09:57 AM
Paps
OH thank you very very much

i couldnt find that equasion, actually i didnt really understand what this wants from me

again really appreciate it :cool: