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7. (a) Use the binomial expansion, in ascending powers of $x$, to show that $$ ext{ } ext{ } \sqrt{4 - x} = 2 - \frac{1}{4}x + kx^2 + .. - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 2

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7.-(a)-Use-the-binomial-expansion,-in-ascending-powers-of-$x$,-to-show-that-$$--ext{-}--ext{-}-\sqrt{4---x}-=-2---\frac{1}{4}x-+-kx^2-+-..-Edexcel-A-Level Maths Pure-Question 8-2017-Paper 2.png

7. (a) Use the binomial expansion, in ascending powers of $x$, to show that $$ ext{ } ext{ } \sqrt{4 - x} = 2 - \frac{1}{4}x + kx^2 + ... $$ where $k$ is a rationa... show full transcript

Worked Solution & Example Answer:7. (a) Use the binomial expansion, in ascending powers of $x$, to show that $$ ext{ } ext{ } \sqrt{4 - x} = 2 - \frac{1}{4}x + kx^2 + .. - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 2

Step 1

Use the binomial expansion to show that $\sqrt{4 - x} = 2 - \frac{1}{4}x + kx^2 + ...$

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Answer

To begin the binomial expansion of 4x\sqrt{4 - x}, we first take out a factor of 4:

4x=4(1x4)=21x4.\sqrt{4 - x} = \sqrt{4(1 - \frac{x}{4})} = 2\sqrt{1 - \frac{x}{4}}.

Now using the binomial expansion for 1+u\sqrt{1 + u} where u=x4u = -\frac{x}{4}, we have:

1x4=1+12(x4)+12(121)12!(x4)2+...\sqrt{1 - \frac{x}{4}} = 1 + \frac{1}{2}\left(-\frac{x}{4}\right) + \frac{1}{2}\left(\frac{1}{2} - 1\right)\frac{1}{2!}\left(-\frac{x}{4}\right)^2 + ...

Calculating each term:

  1. The first term: 11
  2. The second term: 18x-\frac{1}{8}x
  3. The third term: +1128x2+\frac{1}{128}x^2

Thus combining these, we have:

4x=2(118x+1128x2+...)=214x+kx2+...\sqrt{4 - x} = 2\left(1 - \frac{1}{8}x + \frac{1}{128}x^2 + ...\right) = 2 - \frac{1}{4}x + kx^2 + ...

From this, we can see that k=164k = \frac{1}{64}.

Step 2

State, giving a reason, if the expansion is valid for this value of $x$.

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Answer

The expansion is valid for x<4|x| < 4, so x=1x = 1 can be used. This is because the binomial expansion converges within the radius of convergence, which in this case is defined by the condition that the term x4<1|\frac{x}{4}| < 1.

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