3. (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10}), where a is a non-zero constant. Give each term in its simplest form. Given that, in this expansion, the coefficient of x^{3} is double the coefficient of x^{2}, (b) find the value of a.

A-Level Edexcel Maths Pure Equation of a Straight Line: 3. (a) Find the first 4 terms, i

3. (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10}), where a is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 2

Question 5

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3. (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10}), where a is a non-zero constant. Give each term in its simplest... show full transcript

Worked Solution & Example Answer:3. (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10}), where a is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 2

Step 1

Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10})

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Answer

To find the first four terms of the binomial expansion, we can use the Binomial Theorem, which states:

$(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k$

In this case, we have:

$a = 1$
$b = ax$
$n = 10$

Applying the theorem:

For k = 0: ${10 \choose 0} (1)^{10} (ax)^0 = 1$
For k = 1: ${10 \choose 1} (1)^{9} (ax)^1 = 10(ax) = 10ax$
For k = 2: ${10 \choose 2} (1)^{8} (ax)^2 = 45(a^2 x^2) = 45a^2x^2$
For k = 3: ${10 \choose 3} (1)^{7} (ax)^3 = 120(a^3 x^3) = 120a^3 x^3$

Thus, the first four terms in ascending powers of x are:

$1 + 10ax + 45a^2x^2 + 120a^3x^3$

Step 2

find the value of a

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Answer

According to the question, the coefficient of $x^{3}$ is double the coefficient of $x^{2}$ in the expansion.

From our earlier findings:

The coefficient of $x^{2}$ is $45a^2$
The coefficient of $x^{3}$ is $120a^3$

Setting up the equation based on the given condition:

$120a^3 = 2(45a^2)$

Simplifying:

$120a^3 = 90a^2$

Dividing both sides by $a^2$ (since a is non-zero):

$120a = 90$

Solving for a:

$a = \frac{90}{120} = \frac{3}{4}$

Thus, the value of a is:

$a = \frac{3}{4}$

3. (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10}), where a is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 2

Worked Solution & Example Answer:3. (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10}), where a is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 2

Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + (ax)^{10})

find the value of a

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