Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\(
\left(3 - \frac{1}{3} x\right)^5
\)
giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 2
Question 3
Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\(
\left(3 - \frac{1}{3} x\right)^5
\)
giving each term in its simplest form.
Worked Solution & Example Answer:Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\(
\left(3 - \frac{1}{3} x\right)^5
\)
giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 2
Step 1
Finding the First Term
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Answer
The first term of the expansion can be found using the binomial theorem, which states that:
[
(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k
]
In this case, let (a = 3) and (b = -\frac{1}{3}x), with (n = 5). The first term corresponds to (k = 0):
[
{5 \choose 0} (3)^{5} \left(-\frac{1}{3}x\right)^0 = 1 \cdot 243 = 243
]
Step 2
Finding the Second Term
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Answer
The second term corresponds to (k = 1):
[
{5 \choose 1} (3)^{4} \left(-\frac{1}{3}x\right)^1 = 5 \cdot 81 \cdot \left(-\frac{1}{3}x\right) = -135x
]
Step 3
Finding the Third Term
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Answer
The third term corresponds to (k = 2):
[
{5 \choose 2} (3)^{3} \left(-\frac{1}{3}x\right)^2 = 10 \cdot 27 \cdot \left(\frac{1}{9}x^2\right) = 30x^2
]
Step 4
Finding the Fourth Term
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Answer
The fourth term corresponds to (k = 3):
[
{5 \choose 3} (3)^{2} \left(-\frac{1}{3}x\right)^3 = 10 \cdot 9 \cdot \left(-\frac{1}{27}x^3\right) = - rac{10}{3}x^3
]
