Binomial Theorem

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August 13th 2006, 06:34 AM
c00ky

Binomial Theorem

Hi again guys,

Sorry to bother you again, i have another question which i am totally unfamiliar with. If anybody could guide me through the solution it would be great.

Ok here goes...

Questions:

A) Using the technique of the Sum to infinity express as a fraction in its lowest terms the following recurring decimal number

0.05 (recurring)

B) Using the Binomial Theorem find the first 5 terms of the Binomial expression

(1 + 0.2)^12

??? GULP ???
August 13th 2006, 07:07 AM
galactus

The binomial thereom:

$(1+x)^{m}=1+mx+\frac{m(m-1)}{2!}x^{2}+$ $......+\frac{m(m-1)(m-2).......(m-k+1)}{k!}x^{k}$

For yours:

$1+12x+\frac{12(11)}{2!}x^{2}+\frac{12(11)(10)}{3!} x^{3}+....etc$

Then use x=.02
August 13th 2006, 07:11 AM
c00ky

Ok part B i got 8.84544 after i used the theorem.

Can anybody help with part A?
August 13th 2006, 07:15 AM
ThePerfectHacker

Quote:

Originally Posted by c00ky

A) Using the technique of the Sum to infinity express as a fraction in its lowest terms the following recurring decimal number

0.05 (recurring)

I presume you mean,
$.0505050505050...$ .
Then, by definition of decimals,
$\frac{5}{10^2}+\frac{5}{10^4}+\frac{5}{10^6}+...$
Factor, (you can manipulate this series anyway you like since it is absolutely convergent)
$\frac{5}{10^2}\left( 1+\frac{1}{10^2}+\frac{1}{10^4}+... \right)$
Express as sum,
$\frac{5}{100}\sum_{k=0}^{\infty} (100)^{-k}$
Use geometry sum,
$\frac{5}{100}\cdot \frac{1}{1-\frac{1}{100}}$
Simplify,
$\frac{5}{100}\cdot \frac{100}{100-1}=\frac{5}{99}$
August 13th 2006, 07:24 AM
Soroban

Hello, c00ky!

Here's part (A) . . .

Quote:

A) Using the technique of the Sum to infinity, express as a fraction in its lowest terms
the following recurring decimal number: .0.05 (recurring)

It's not clear what part is recurring.
. . I'll assume that the number is: . $N \:= \:0.0555...$

We have: . $N\;=\;\frac{5}{100} + \frac{5}{1000} + \frac{5}{10,000} + \frac{5}{100,000} + \hdots$

Factor: . . $N \;= \;\frac{5}{100}\underbrace{\left(1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \hdots\right)}$

We have a geometric series with $a = 1,\;r = \frac{1}{10}$
. . Its sum is: . $S\:=\:\frac{1}{1 - \frac{1}{10}} \:=\:\frac{10}{9}$

Therefore: . $N\;=\;\left(\frac{5}{100}\right)\left(\frac{10}{9} \right)\;= \;\boxed{\frac{1}{18}}$
August 13th 2006, 07:31 AM
c00ky

A BIG thanks to everybody. Soroban that's the part which is recurring, that's great thanks.

:)