A-Level Edexcel Maths: Pure Differentiation: Use the binomial series to find

Use the binomial series to find the expansion of $$ \frac{1}{(2 + 5x)^3}, \\ |x| < \frac{2}{5} $$ in ascending powers of $x$, up to and including the term in $x^3$ - Edexcel - A-Level Maths: Pure - Question 3 - 2016 - Paper 4

Question 3

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Use the binomial series to find the expansion of $$ \frac{1}{(2 + 5x)^3}, \\ |x| < \frac{2}{5} $$ in ascending powers of $x$, up to and including the term in $x^3$.... show full transcript

Worked Solution & Example Answer:Use the binomial series to find the expansion of $$ \frac{1}{(2 + 5x)^3}, \\ |x| < \frac{2}{5} $$ in ascending powers of $x$, up to and including the term in $x^3$ - Edexcel - A-Level Maths: Pure - Question 3 - 2016 - Paper 4

Step 1

Step 1: Identify the Binomial Series

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Answer

The binomial series for $(1 + u)^n$ can be expressed as:

(1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k

for $|u| < 1$ . We will express $rac{1}{(2 + 5x)^3}$ in this form.

Step 2

Step 2: Rewrite the Expression

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Answer

Start by rewriting the expression:

\frac{1}{(2 + 5x)^3} = \frac{1}{2^3} \cdot \frac{1}{(1 + \frac{5x}{2})^3} = \frac{1}{8} \cdot (1 + \frac{5x}{2})^{-3}

Step 3

Step 3: Apply the Binomial Theorem

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Answer

Now, apply the binomial series:

(1 + u)^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k} u^k = \sum_{k=0}^{\infty} (-1)^k \frac{3(3 + 1)(3 + 2)\ldots (3 + k - 1)}{k!} u^k

Substituting $u = \frac{5x}{2}$ gives:

(1 + \frac{5x}{2})^{-3} = \sum_{k=0}^{\infty} (-1)^k \binom{3+k-1}{k} \left(\frac{5x}{2}\right)^k

Step 4

Step 4: Collect the Terms

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Answer

We need to collect terms up to $x^3$ :

For $k=0$ : $inom{3}{0}(\frac{5}{2})^0 = 1$
For $k=1$ : $inom{4}{1}(\frac{5}{2})^1 = 20(\frac{5}{2}) = 50$
For $k=2$ : $inom{5}{2}(\frac{5}{2})^2 = 10(\frac{25}{4}) = \frac{250}{4} = 62.5$
For $k=3$ : $inom{6}{3}(\frac{5}{2})^3 = 20(\frac{125}{8}) = \frac{2500}{8} = 312.5$ Hence, combining these terms, we have:

\frac{1}{8} (1 - 50x + 62.5x^2 - 312.5x^3)

Step 5

Step 5: Final Expansion

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Answer

Multiplying by $rac{1}{8}$ gives:

\frac{1}{8} - \frac{50}{8}x + \frac{62.5}{8}x^2 - \frac{312.5}{8}x^3

Thus, the expansion up to $x^3$ is:

\frac{1}{8} - \frac{25}{4}x + \frac{15.625}{4}x^2 - \frac{39.0625}{4}x^3

With coefficients in their simplest form, we have:

Use the binomial series to find the expansion of $$ \frac{1}{(2 + 5x)^3}, \\ |x| < \frac{2}{5} $$ in ascending powers of $x$, up to and including the term in $x^3$ - Edexcel - A-Level Maths: Pure - Question 3 - 2016 - Paper 4

Worked Solution & Example Answer:Use the binomial series to find the expansion of $$ \frac{1}{(2 + 5x)^3}, \\ |x| < \frac{2}{5} $$ in ascending powers of $x$, up to and including the term in $x^3$ - Edexcel - A-Level Maths: Pure - Question 3 - 2016 - Paper 4

Step 1: Identify the Binomial Series

Step 2: Rewrite the Expression

Step 3: Apply the Binomial Theorem

Step 4: Collect the Terms

Step 5: Final Expansion

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