# Math Help - Binomial Expension

1. ## Binomial Expension

Write down the first three term in the binomial expansion of $3\sqrt {1 + \frac{x}{125}}$.Hence,deduce that $3\sqrt {126} \approx 5.013$.

2. ## Re: Binomial Expension

Originally Posted by mastermin346
Write down the first three term in the binomial expansion of $3\sqrt {1 + \frac{x}{125}}$.Hence,deduce that $3\sqrt {126} \approx 5.013$.
Write this as $\displaystyle 3\left(1 + \frac{x}{125}\right)^{\frac{1}{2}}$.

You should know that the binomial series is $\displaystyle (1 + X)^{\alpha} = 1 + \alpha X + \frac{\alpha(\alpha - 1)}{2!}X^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!}X^3 + \dots$. Here you need to let $\displaystyle X = \frac{x}{125}$ and $\displaystyle \alpha = \frac{1}{2}$.

3. ## Re: Binomial Expension

$3\left ( 1 + \frac{x}{125} \right )^\frac{1}{2}$ let
$a = 1$
$b = \frac{x}{125}$ and $n = \frac{1}{2}$

I get $3 + \frac{3}{250}x - \frac{3}{125000}x^2 + \frac{3}{31250000}x^3 + ...$

Then?

4. ## Re: Binomial Expension

Originally Posted by mastermin346
$3\left ( 1 + \frac{x}{125} \right )^\frac{1}{2}$ let
$a = 1$
$b = \frac{x}{125}$ and $n = \frac{1}{2}$

I get $3 + \frac{3}{250}x - \frac{3}{125000}x^2 + \frac{3}{31250000}x^3 + ...$

Then?
Correct.

Now note that $\displaystyle 3\sqrt{126} = 3\sqrt{1 + 125} = 3\sqrt{1 + \frac{15625}{125}}$. So what do you need to let $\displaystyle x$ equal in your series?

5. ## Re: Binomial Expension

$x = 15625$

6. ## Re: Binomial Expension

Originally Posted by mastermin346
Write down the first three term in the binomial expansion of $\sqrt[3] {1 + \frac{x}{125}}$.Hence,deduce that $\sqrt[3] {126} \approx 5.013$.
Clearly the question should be $\color{blue}\sqrt[3] {126}$

7. ## Re: Binomial Expension

Originally Posted by Plato
Clearly the question should be $\color{blue}\sqrt[3] {126}$
Yes that would probably make more sense :P

8. ## Re: Binomial Expension

One has to represent $\sqrt[3]{126}$ as $5\sqrt[3]{1+\frac{1}{125}}$. Taylor expansion for $\sqrt[3]{1+x}$ works well when $x\approx 0$. One can check the Taylor expansion

$(1+x)^{1/3}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{5x^3}{81}+\dots$

for x = 125 gives 118869.8, which is very far from the correct value 5.013. On the other hand, the Lagrange form of the remainder says that $(1+x)^{1/3}=1+\frac{x}{3}+R_2(x)$ where $|R_2(1/125)|<10^{-5}$, so $5(1+x)^{1/3}=5+\frac{5x}{3}+5R_2(x)$ and $|5R_2(1/125)|<10^{-4}$.