A-Level Edexcel Maths Pure Graphs of Functions: 3. (a) Find the first four terms

3. (a) Find the first four terms, in ascending powers of x, in the binomial expansion of (1+kx)^6, where k is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Question 5

3.-(a)-Find-the-first-four-terms,-in-ascending-powers-of-x,-in-the-binomial-expansion-of--(1+kx)^6,-where-k-is-a-non-zero-constant-Edexcel-A-Level Maths Pure-Question 5-2007-Paper 2.png

3. (a) Find the first four terms, in ascending powers of x, in the binomial expansion of (1+kx)^6, where k is a non-zero constant. Given that, in this expansion, ... show full transcript

Worked Solution & Example Answer:3. (a) Find the first four terms, in ascending powers of x, in the binomial expansion of (1+kx)^6, where k is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Step 1

Find the first four terms in ascending powers of x

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Answer

The binomial expansion can be expressed as:

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

For our scenario:

$a = 1$ , $b = kx$ , and $n = 6$ .
The first four terms (for $r = 0, 1, 2, 3$ ) are computed as follows:

When $r=0$ :
${{6 \choose 0}} (1)^{6} (kx)^{0} = 1$
When $r=1$ :
${{6 \choose 1}} (1)^{5} (kx)^{1} = 6kx$
When $r=2$ :
${{6 \choose 2}} (1)^{4} (kx)^{2} = 15k^2 x^2$
When $r=3$ :
${{6 \choose 3}} (1)^{3} (kx)^{3} = 20k^3 x^3$

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

$1 + 6kx + 15k^2 x^2 + 20k^3 x^3$

Step 2

the value of k

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Answer

To find the value of $k$ , we set the coefficients of $x$ and $x^2$ equal:

From the previous step:

Coefficient of $x$ : $6k$
Coefficient of $x^2$ : $15k^2$

Setting them equal gives: $6k = 15k^2$

Rearranging yields: $15k^2 - 6k = 0$ Factoring: $k(15k - 6) = 0$

Since $k$ is non-zero, $15k - 6 = 0 \Rightarrow k = \frac{6}{15} = \frac{2}{5}$

Step 3

the coefficient of x^3

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Answer

Using the found value of $k = \frac{2}{5}$ , we find the coefficient of $x^3$ :

From the first four terms:

Coefficient for $x^3$ : $20k^3$
Substituting for $k$ : $20 \left( \frac{2}{5} \right)^3 = 20 \left( \frac{8}{125} \right) = \frac{160}{125} = \frac{32}{25}$

Thus, the coefficient of $x^3$ is $\frac{32}{25}$ .

3. (a) Find the first four terms, in ascending powers of x, in the binomial expansion of (1+kx)^6, where k is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Worked Solution & Example Answer:3. (a) Find the first four terms, in ascending powers of x, in the binomial expansion of (1+kx)^6, where k is a non-zero constant - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 2

Find the first four terms in ascending powers of x

the value of k

the coefficient of x^3

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