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(a) Find the binomial expansion of $(4 + 5x)^{\frac{1}{2}}$, $ |x| < \frac{4}{5} $ give each coefficient in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 4

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(a)-Find-the-binomial-expansion-of---$(4-+-5x)^{\frac{1}{2}}$,---$-|x|-<-\frac{4}{5}-$---give-each-coefficient-in-its-simplest-form-Edexcel-A-Level Maths Pure-Question 2-2015-Paper 4.png

(a) Find the binomial expansion of $(4 + 5x)^{\frac{1}{2}}$, $ |x| < \frac{4}{5} $ give each coefficient in its simplest form. (b) Find the exact value of $(4... show full transcript

Worked Solution & Example Answer:(a) Find the binomial expansion of $(4 + 5x)^{\frac{1}{2}}$, $ |x| < \frac{4}{5} $ give each coefficient in its simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 4

Step 1

Find the binomial expansion of $(4 + 5x)^{\frac{1}{2}}$

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Answer

To find the binomial expansion of (4+5x)12(4 + 5x)^{\frac{1}{2}}, we can use the binomial theorem, which states:

(a+b)n=k=0n(nk)ankbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

In this case, let:

  • a=4a = 4
  • b=5xb = 5x
  • n=12n = \frac{1}{2}

Thus, we get:

(4+5x)12=k=0n(12k)412k(5x)k(4 + 5x)^{\frac{1}{2}} = \sum_{k=0}^{n} \binom{\frac{1}{2}}{k} 4^{\frac{1}{2}-k} (5x)^k

Calculating the first few terms:

  • For k=0k=0: (inom{\frac{1}{2}}{0} 4^{\frac{1}{2}} (5x)^0 = 2)
  • For k=1k=1: (inom{\frac{1}{2}}{1} 4^{-\frac{1}{2}} (5x)^1 = \frac{1}{2} \cdot \frac{1}{2} \cdot 5x = \frac{5}{4}x)
  • For k=2k=2: (inom{\frac{1}{2}}{2} 4^{-\frac{3}{2}} (5x)^2 = \frac{1}{2} \cdot -\frac{1}{2} \cdot \frac{1}{12} \cdot (25x^2) = -\frac{25}{48}x^2)

Thus, the expansion up to x2x^2 is:

2+54x2548x22 + \frac{5}{4} x - \frac{25}{48} x^2.

Step 2

Find the exact value of $(4 + 5x)^{\frac{1}{2}}$ when $x = \frac{1}{10}$

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Answer

Substituting x=110x = \frac{1}{10} into the expression derived in part (a):

(4+5110)12=(4+52)12=(132)12=132=k2(4 + 5 \cdot \frac{1}{10})^{\frac{1}{2}} = (4 + \frac{5}{2})^{\frac{1}{2}} = (\frac{13}{2})^{\frac{1}{2}} = \frac{\sqrt{13}}{\sqrt{2}} = k\sqrt{2}

We can express 13=k2\sqrt{13} = k\sqrt{2} where k=132k = \frac{\sqrt{13}}{\sqrt{2}}.

Step 3

Substitute $x = \frac{1}{10}$ into your binomial expansion and find an approximate value for $\sqrt{2}$

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Answer

Substituting x=110x = \frac{1}{10} into the binomial expansion:

2+541102548(110)22 + \frac{5}{4} \cdot \frac{1}{10} - \frac{25}{48} \cdot \left(\frac{1}{10}\right)^2

Calculating each part:

  • First term: 22
  • Second term: 54110=540=18\frac{5}{4} \cdot \frac{1}{10} = \frac{5}{40} = \frac{1}{8}
  • Third term: 25481100=254800- \frac{25}{48} \cdot \frac{1}{100} = -\frac{25}{4800}

Adding these together: Approximation =2+18254800\text{Approximation } = 2 + \frac{1}{8} - \frac{25}{4800}

Converting 22 to a common denominator gives: 96004800+6004800254800=9600+600254800=101754800\frac{9600}{4800} + \frac{600}{4800} - \frac{25}{4800} = \frac{9600 + 600 - 25}{4800} = \frac{10175}{4800}

Thus, the approximate value of 2\sqrt{2} can be expressed as: 101754800\frac{10175}{4800}

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