\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Number and Algebra

Syllabus Content

The binomial theorem including the expansion of (a+b)n,n ∈ N. Use of Pascal's triangle and nCr

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Official Guidance, clarification and syllabus links:

Counting principles may be used in the development of the theorem.

\(^nC_r\) should be found using both the formula and technology.

Example: Find \(r\) when \(^6C_r=20\), using a table of values generated with technology.

Formula Booklet:

Binomial theorem \(n \in \mathbb{N} \)

$$ (a + b)^n = a^n +\;^nC_1 a^{n-1}b + \ldots + \;^nC_r a^{n-r}b^r + \ldots + b^n $$ $$ ^nC_r = \frac{n!}{r!(n-r)!} $$


To expand \( (2x + 3y)^5 \) using the binomial theorem, we can use the formula:

$$ (a + b)^n = a^n + nC_1 a^{n-1}b + \ldots + nC_r a^{n-r}b^r + \ldots + b^n $$

Substituting \( a = 2x \), \( b = 3y \), and \( n = 5 \) into the formula, we get:

$$ (2x + 3y)^5 = (2x)^5 + 5C_1 (2x)^4(3y) + 5C_2 (2x)^3(3y)^2 + 5C_3 (2x)^2(3y)^3 + 5C_4 (2x)(3y)^4 + (3y)^5 $$

Simplifying further:

$$ (2x + 3y)^5 = 32x^5 + 160x^4 \cdot 3y + 80x^3 \cdot 9y^2 + 40x^2 \cdot 27y^3 + 10x \cdot 81y^4 + 243y^5 $$

Which results in:

$$ (2x + 3y)^5 = 32x^5 + 480x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5 $$

This video on the Binomial Theorem is from Revision Village and is aimed at students taking the IB Maths AA Standard level course

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