AskSia

Plus

1. (a) Find the first 4 terms, in ascending powers of

x

, of the binomial expan...

Mar 15, 2024

1. (a) Find the first 4 terms, in ascending powers of

x

, of the binomial expansion of \left(\frac{1}{4}-5 x\right)^{\frac{1}{2}} \quad|x|<\frac{1}{20} giving each coefficient in its simplest form. By substituting

x=\frac{1}{100}

into the answer for (a), (b) find an approximation for

\sqrt{5}

Give your answer in the form

\frac{a}{b}

where

a

and

b

are integers to be found. (a)

\left(\frac{1}{4}\right)^{\frac{1}{2}}\left[1-\frac{5}{4} x\right)^{\frac{1}{2}}

=

Solution by Steps

step 1

To find the first 4 terms of the binomial expansion of

\left(\frac{1}{4}-5x\right)^{\frac{1}{2}}

, we use the Taylor series expansion around

x=0

up to the

x^4

term

step 2

The Taylor series expansion provided by asksia-ll calculator is

1/2 - 5x - 25x^2 - 250x^3 - 3125x^4 + O(x^5)

step 3

The first 4 terms in ascending powers of

x

are:

1/2

-5x

-25x^2

, and

-250x^3

step 4

By substituting

x=\frac{1}{100}

into the expansion, we approximate

\sqrt{5}

\sqrt{\frac{1}{4}-5\left(\frac{1}{100}\right)}

step 5

Substituting

x=\frac{1}{100}

into the first 4 terms gives us

\frac{1}{2} - 5\left(\frac{1}{100}\right) - 25\left(\frac{1}{100}\right)^2 - 250\left(\frac{1}{100}\right)^3

step 6

Simplifying the expression from step 5, we get

\frac{1}{2} - \frac{5}{100} - \frac{25}{10000} - \frac{250}{1000000}

step 7

Further simplifying, we get

\frac{1}{2} - \frac{1}{20} - \frac{1}{400} - \frac{1}{4000}

step 8

Adding the terms together, we get

\frac{1}{2} - \frac{1}{20} - \frac{1}{400} - \frac{1}{4000} = \frac{2000}{4000} - \frac{200}{4000} - \frac{10}{4000} - \frac{1}{4000} = \frac{1789}{4000}

step 9

Therefore, an approximation for

\sqrt{5}

\frac{1789}{4000}

Answer

\frac{1789}{4000}

Key Concept

Binomial Expansion and Substitution

Explanation

The binomial expansion is used to express

(\frac{1}{4}-5x)^{\frac{1}{2}}

in a series form. By substituting

x=\frac{1}{100}

into the expansion, we can approximate

\sqrt{5}