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 Circles: 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

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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 f... 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 of

$(1 + ax)^{10},$

we can apply the Binomial Theorem:

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

Here,\n- $n = 10$ , $a = 1$ , and $b = ax$ . Thus, we have:

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

Combining these, the first four terms are:

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

Step 2

Given that, in this expansion, the coefficient of x^3 is double the coefficient of x^2, find the value of a.

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Answer

From the expansion:

Coefficient of $x^3$ is $120a^3$
Coefficient of $x^2$ is $45a^2$

According to the problem statement, we can express this relationship as:

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

This simplifies to:

$120a^3 = 90a^2$

Dividing both sides by $a^2$ (assuming $a \neq 0$ ):

$120a = 90$

Solving for $a$ gives us:

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

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}

Given that, in this expansion, the coefficient of x^3 is double the coefficient of x^2, find the value of a.

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