A-Level Edexcel Maths Pure Modelling with 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 ex... 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

To find the first four terms of the binomial expansion of $(1 + kx)^6$ , we use the binomial theorem:

$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ where $a = 1$ , $b = kx$ , and $n = 6.$

Calculating the first four terms:

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

Thus, the first four terms are: $1 + 6kx + 15k^2x^2 + 20k^3x^3$

Step 2

the value of $k$

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Answer

To find the value of $k$ , we need to equate the coefficients of $x$ and $x^2$ :

From the expansion:

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

Setting them equal: $6k = 15k^2$ Rearranging this gives: $15k^2 - 6k = 0$ Factoring out $k$ :
$k(15k - 6) = 0$ Thus, $k = 0$ or $k = \frac{6}{15} = \frac{2}{5}$ . Since $k$ is a non-zero constant, we have: $k = \frac{2}{5}$

Step 3

the coefficient of $x^3$

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Answer

Now that we have the value of $k = \frac{2}{5}$ , we can find the coefficient of $x^3$ :

From the earlier expansion, the coefficient of $x^3$ was: $20k^3$ Substituting $k$ : $20 \left( \frac{2}{5} \right)^3 = 20 \times \frac{8}{125} = \frac{160}{125} = \frac{32}{25}$

Therefore, 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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