4. (a) Write down the first three terms, in ascending powers of x, of the binomial expansion of (1 + px²), where p is a non-zero constant. Given that, in the expansion of (1 + px²)¹², the coefficient of x is -q and the coefficient of x² is 11q. (b) find the value of p and the value of q.

A-Level Edexcel Maths Pure Modelling with Functions: 4. (a) Write down the first thre

4. (a) Write down the first three terms, in ascending powers of x, of the binomial expansion of (1 + px²), where p is a non-zero constant - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2

Question 6

4.-(a)-Write-down-the-first-three-terms,-in-ascending-powers-of-x,-of-the-binomial-expansion-of-(1-+-px²),-where-p-is-a-non-zero-constant-Edexcel-A-Level Maths Pure-Question 6-2005-Paper 2.png

4. (a) Write down the first three terms, in ascending powers of x, of the binomial expansion of (1 + px²), where p is a non-zero constant. Given that, in the expans... show full transcript

Worked Solution & Example Answer:4. (a) Write down the first three terms, in ascending powers of x, of the binomial expansion of (1 + px²), where p is a non-zero constant - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2

Step 1

(a) Write down the first three terms, in ascending powers of x

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Answer

To find the first three terms of the binomial expansion of $(1 + px^2)^{12}$ , we can use the binomial theorem:

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

In our case, let $a = 1$ , $b = px^2$ , and $n = 12$ :

First term (k=0): $\binom{12}{0} (1)^{12} (px^2)^0 = 1$
Second term (k=1): $\binom{12}{1} (1)^{11} (px^2)^1 = 12px^2$
Third term (k=2): $\binom{12}{2} (1)^{10} (px^2)^2 = \frac{12 \times 11}{2} p^2 x^4 = 66p^2 x^4$

Thus, the first three terms are:

$1 + 12px^2 + 66p^2 x^4$

Step 2

(b) find the value of p and the value of q.

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Answer

From the given information:

The coefficient of $x$ is $-q$ , which corresponds to the coefficient of the $x$ term in the expansion.
The coefficient of $x^2$ is $11q$ .

To find these coefficients, note:

Coefficient of $x$ : It comes from the term with $k=1$ . Thus, $12p = -q$
Coefficient of $x^2$ : It comes from the term with $k=2$ . Thus, $66p^2 = 11q$

Now, substituting $q$ from the first equation into the second:

Substitute $q$ : $66p^2 = 11(-12p)$ $66p^2 = -132p$
Rearranging gives: $66p^2 + 132p = 0$ $p(66p + 132) = 0$

Since $p$ is non-zero, we divide: $66p + 132 = 0 \Rightarrow p = -2$

Substitute $p$ back into the equation for $q$ : $q = -12p = -12(-2) = 24$

Thus, the values are:

$p = -2$
$q = 24$

4. (a) Write down the first three terms, in ascending powers of x, of the binomial expansion of (1 + px²), where p is a non-zero constant - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2

Worked Solution & Example Answer:4. (a) Write down the first three terms, in ascending powers of x, of the binomial expansion of (1 + px²), where p is a non-zero constant - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2

(a) Write down the first three terms, in ascending powers of x

(b) find the value of p and the value of q.

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