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2. (a) Use the binomial theorem to find all the terms of the expansion of $(2 + 3x)^4$ Give each term in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 4

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2.-(a)-Use-the-binomial-theorem-to-find-all-the-terms-of-the-expansion-of--$(2-+-3x)^4$-Give-each-term-in-its-simplest-form-Edexcel-A-Level Maths Pure-Question 4-2013-Paper 4.png

2. (a) Use the binomial theorem to find all the terms of the expansion of $(2 + 3x)^4$ Give each term in its simplest form. (b) Write down the expansion of $(2 ... show full transcript

Worked Solution & Example Answer:2. (a) Use the binomial theorem to find all the terms of the expansion of $(2 + 3x)^4$ Give each term in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 4

Step 1

Use the binomial theorem to find all the terms of the expansion of $(2 + 3x)^4$

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Answer

To find the expansion of (2+3x)4(2 + 3x)^4, we apply the binomial theorem which states:

(a + b)^n = inom{n}{0}a^n b^0 + inom{n}{1}a^{n-1}b^1 + inom{n}{2}a^{n-2}b^2 + ... + inom{n}{n}a^0b^n

Here, we have:

  • a=2a = 2
  • b=3xb = 3x
  • n=4n = 4

Using the theorem, we get:

  1. For k=0k = 0: inom{4}{0} (2)^{4} (3x)^0 = 1 imes 16 = 16

  2. For k=1k = 1: inom{4}{1} (2)^{3} (3x)^{1} = 4 imes 8 imes 3x = 96x

  3. For k=2k = 2: inom{4}{2} (2)^{2} (3x)^{2} = 6 imes 4 imes 9x^2 = 216x^2

  4. For k=3k = 3: inom{4}{3} (2)^{1} (3x)^{3} = 4 imes 2 imes 27x^3 = 216x^3

  5. For k=4k = 4: inom{4}{4} (2)^{0} (3x)^{4} = 1 imes 81x^4 = 81x^4

Thus, the complete expansion is: 16+96x+216x2+216x3+81x416 + 96x + 216x^2 + 216x^3 + 81x^4

Therefore, the terms are:

  • 1616
  • 96x96x
  • 216x2216x^2
  • 216x3216x^3
  • 81x481x^4

Step 2

Write down the expansion of $(2 - 3x)^4$ in ascending powers of $x$, giving each term in its simplest form.

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Answer

To expand (23x)4(2 - 3x)^4, we again apply the binomial theorem:

(a - b)^n = inom{n}{0}a^n(-b)^0 + inom{n}{1}a^{n-1}(-b)^1 + inom{n}{2}a^{n-2}(-b)^2 + ... + inom{n}{n}a^0(-b)^n

In this case:

  • a=2a = 2
  • b=3xb = 3x
  • n=4n = 4

Applying the theorem, we have:

  1. For k=0k = 0: inom{4}{0} (2)^4 (-3x)^0 = 1 imes 16 = 16

  2. For k=1k = 1: inom{4}{1} (2)^{3} (-3x)^{1} = 4 imes 8 imes (-3x) = -96x

  3. For k=2k = 2: inom{4}{2} (2)^{2} (-3x)^{2} = 6 imes 4 imes 9x^2 = 216x^2

  4. For k=3k = 3: inom{4}{3} (2)^{1} (-3x)^{3} = 4 imes 2 imes (-27x^3) = -216x^3

  5. For k=4k = 4: inom{4}{4} (2)^{0} (-3x)^{4} = 1 imes 81x^4 = 81x^4

Thus, the expansion is: 1696x+216x2216x3+81x416 - 96x + 216x^2 - 216x^3 + 81x^4

Therefore, the terms in ascending order of xx are:

  • 1616
  • 96x-96x
  • 216x2216x^2
  • 216x3-216x^3
  • 81x481x^4

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