IB Mathematics AA SL Flashcards- Binomial theorem

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[h] IB Mathematics AA SL Flashcards- Binomial theorem

[q] Binomial theorem

[a]

INVESTIGATION// Expand the following:

\[
(a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2
\]

\[
(a + b)^3 = (a + b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3
\]

[q]

SITUATION// In this section, we will be dealing with any kind of expansion in the form: \((a + b)^n\), such as \((2x + 3)^2\) or \(\left(x^3 – \frac{2}{x^2}\right)^8\). As you can see above, we will have specific coefficients that multiply each term, shown in orange. It will depend on the power \((n)\), and which powers of \(a\) & \(b\) you are considering.

[a]

COEFFICIENTS// Let’s consider how we found the ‘3ab^2’. This was all the different ways of choosing two b’s from the three brackets: \((a + b)(a + b)(a + b)\).

For any situation, if we call the power of \(b\), ‘r’—then this coefficient will be equal to the number of ways to choose \(r\) objects from \(n\) total objects.

[q]

‘n CHOOSE r’// This is a well-established branch of maths, and we have a few ways of calculating these coefficients. The notation is \(\binom{n}{r}\) or nCr.

[a]

1// PASCAL’S TRIANGLE// In this seemingly unrelated pattern, where every number is found by adding the two numbers above, each set of values for \(n\) is represented by each row. Then, the leftmost value is for \(r = 0\), and as you move right, \(r\) increases.

[q]

2→ FORMULA// This may be revisited in 1.10, but nCr can simply be calculated using the formula:

E.G.1 // Calculate \(\binom{7}{5}\):

Using the formula…

\[
\binom{7}{5} \text{ or } 7C5 = \frac{7!}{5! \cdot (7-5)!} = \frac{7 \times 6 \times 5!}{5! \cdot 2!} = \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21
\]

[a]

3→ GDC// The ‘nCr’ function can be done on your GDC, which is helpful if the numbers start getting quite big.

\(\text{TI-nspire} \rightarrow \text{MENU} \rightarrow 5: \text{PROB} \rightarrow 3: \text{COMB} \rightarrow \text{TYPE: nCr}(n,r) \rightarrow \text{ENTER}\)

\(\text{TI-84} \rightarrow \text{TYPE:} \rightarrow \text{MATH} \rightarrow \text{PRB} \rightarrow 3: \text{nCr} \rightarrow \text{TYPE: nCr}(n,r) \rightarrow \text{ENTER}\)

[q]

1// Now that we know how to calculate the nCr coefficients, we can put together the whole expansion, with descending powers of \(a\) & ascending powers of \(b\):

\[
(a+b)^n = a^n + \binom{n}{1}a^{n-1}b + \ldots + \binom{r}{r}a^{n-r}b^r + \ldots + b^n
\]

[a]

1// The common questions ask you to find one specific term out of the expansion, such as the ‘x\(^2\) term’ or the ‘constant term’. The final coefficient will also include powers of constants included in ‘a’ or ‘b’.

[q]

E.G.1 // Fully expand \((2x+3)^4\):

Using the formula, we get:

\[
(2x)^4 + \binom{4}{1}(2x)^3(3) + \binom{4}{2}(2x)^2(3)^2 + \binom{4}{3}(2x)(3)^3 + (3)^4
\]

This gives:

\[
16x^4 + 4 \times 8x^3 \times 3 + 6 \times 4x^2 \times 9 + 4 \times 2x \times 27 + 81 = 16x^4 + 96x^3 + 216x^2 + 216x + 81
\]

[a]

E.G.2 // Find the \(x^3\) term in the expansion of \((2x^2 – \frac{1}{x})^6\):

You could try to observe the pattern of the exponents, which would be \(x^{12}, x^6, x^0, \ldots\), which would mean we should find the fourth term. Or, using algebra, the general term is \(\binom{6}{r}a^{n-r}b^r\), so

\[
\binom{6}{3}(2x^2)^{6-3}\left(-\frac{1}{x}\right)^3
\]

[q]

Ignoring the coefficients for a second, we have

\[
x^{12-3r}\left(-x^{1-r}\right) = x^{12-3r}x^{-r} = x^{12-4r}
\]

As we need the \(x^3\) term, we set \(12-4r=3\) \(\Rightarrow\) \(r=3\). Then we evaluate the term:

\[
\binom{6}{3}(2x^2)^{6-3}\left(-\frac{1}{x}\right)^3 = 20x^6(-4x^{-3}) = -10240x^3
\]

[a]

E.G.3 // (IB 2014) The expansion of \(\left(\frac{x^3}{2} + \frac{p}{x}\right)^8\) has constant term: 5103. Find possible values of \(p\):

Even though this one is reversed, we will still start by finding which term has the correct power of \(x \rightarrow x^0\).

[q]

Using the algebraic method, ignoring coefficients, we see that the general term has the form:

\[
x^{3(8-r)}r \times p^r x^c
\]

[a]

meaning we solve \(x^{24-4r}=0\), so \(r=6\). We then plug that, along with \(n=8\), \(a=\frac{x^3}{2}\), \(b=\frac{p}{x}\) into the formula, and set it equal to 5103:

\[
\binom{8}{6}\left(\frac{x^3}{2}\right)^{2} \times p^2 \times x^c = 5103 \Rightarrow 28 \times \frac{1}{4} \times p^6 \times \frac{x^c}{x^c} = 5103 \Rightarrow 7p^6 = 5103
\]

\[
p^6 = 729 \Rightarrow p = \sqrt[6]{729} \Rightarrow p = \pm 3
\]

[a] Standard form:

A number in the format , \( \pm a \times10^k \) , where \(1\leqslant a\leqslant 10 \) and \(k\in Z\) (k is an integer)

(enter text or “Add Media”; select text to format)

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