(a) Write down the first three terms, in ascending powers of $x$, of the binomial expansion of $(1 + px)^n$, where $p$ is a non-zero constant. Given that, in the expansion of $(1 + px)^{12}$, the coefficient of $x$ is $-q$ and the coefficient of $x^2$ is $11q$. (b) find the value of $p$ and the value of $q$.

A-Level Edexcel Maths Pure Integration: (a) Write down the first three t

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

Question 6

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

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

Step 1

Write down the first three terms, in ascending powers of $x$, of the binomial expansion of $(1 + px)^{12}$

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Answer

The first three terms of the binomial expansion can be obtained using the binomial theorem, which states that:

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

In our case, we take $a = 1$ , $b = px$ , and $n = 12$ :

For $k=0$ : $\binom{12}{0}(1)^{12}(px)^0 = 1$
For $k=1$ : $\binom{12}{1}(1)^{11}(px)^1 = 12(px) = 12px$
For $k=2$ : $\binom{12}{2}(1)^{10}(px)^2 = 66(px)^2 = 66p^2x^2$

Thus, the first three terms are:

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

Step 2

Given that, in the expansion of $(1 + px)^{12}$, the coefficient of $x$ is $-q$ and the coefficient of $x^2$ is $11q$; find $p$ and $q$

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Answer

From the earlier expansion, we know:

The coefficient of $x$ is $12p = -q$ ;
The coefficient of $x^2$ is $66p^2 = 11q$ .

We can use the first equation to express $q$ in terms of $p$ :

$q = -12p$

Substituting this into the second equation gives:

$66p^2 = 11(-12p)$ \Rightarrow 66p^2 = -132p $\Rightarrow 66p^2 + 132p = 0\Rightarrow 66p(p + 2) = 0$

This gives us solutions:

$p = 0$ (not valid since $p$ is non-zero), or
$p = -2$ .

Substituting $p = -2$ back into the expression for $q$ :

$q = -12(-2) = 24.$

Thus, the values are:

$p = -2$
$q = 24$

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

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

Write down the first three terms, in ascending powers of $x$, of the binomial expansion of $(1 + px)^{12}$

Given that, in the expansion of $(1 + px)^{12}$, the coefficient of $x$ is $-q$ and the coefficient of $x^2$ is $11q$; find $p$ and $q$

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