How is each number in Pascal's triangle worked out?

Each number is the sum of the two numbers above it . It helps to imagine zeros sitting outside the triangle, which is what produces the 1 at each end of every row.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsBinomial Expansion

Binomial Expansion (AQA GCSE Further Maths): Flashcards

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How is each number in Pascal's triangle worked out?

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How is each number in Pascal's triangle worked out?
Each number is the sum of the two numbers above it.
It helps to imagine zeros sitting outside the triangle, which is what produces the 1 at each end of every row.
One row of Pascal's triangle is 1, 4, 6, 4, 1. Complete the row below it:
$1,______, 10, 10,______, 1$
The completed row is:
$1, 5, 10, 10, 5, 1$
Each entry comes from adding the pair above it, so $1 + 4 = 5$ and $4 + 6 = 10$ . Notice that the row reads the same forwards and backwards, as every row does.
Define binomial.
A binomial is the sum or difference of two different terms, such as $3 + 4 x$ or $p - q$ .
A binomial expansion is what you get when a binomial is raised to a power, ${(a + b)}^{n}$ .
Which row of Pascal's triangle gives the coefficients for ${(a + b)}^{n}$ ?
The row that starts 1, $n$ , so for ${(a + b)}^{3}$ you use the row beginning 1, 3.
That row is 1, 3, 3, 1, which gives the four coefficients of the expansion.
Complete the expansion:
${(a + b)}^{4} = a^{4} + 4 a^{3} b +______+ 4 a b^{3} + b^{4}$
The completed expansion is:
${(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$
The powers of $a$ decrease while the powers of $b$ increase, and in every term the two powers add up to 4.
True or False?
The expansion of ${(a + b)}^{n}$ has $n + 1$ terms.
True.
The powers of $a$ run from $n$ all the way down to 0, and that is $n + 1$ different values.
So ${(a + b)}^{4}$ has five terms, not four.
When expanding ${(2 x + 3)}^{4}$ , why must you put brackets round the $2 x$ ?
Because the power applies to the whole term, so ${(2 x)}^{3} = 2^{3} x^{3} = 8 x^{3}$ rather than $2 x^{3}$ .
Leaving the bracket off loses the powers of the number, and every coefficient then comes out wrong.
The Pascal coefficients for ${(a + b)}^{4}$ are 1, 4, 6, 4, 1. Why is the expansion of ${(2 x + 3)}^{4}$ not symmetric?
The expansion is $16 x^{4} + 96 x^{3} + 216 x^{2} + 216 x + 81$ , whose coefficients do not read the same both ways.
The Pascal numbers are symmetric, but each one is then multiplied by different powers of 2 and 3, and that destroys the symmetry.
What happens to the signs when you expand ${(2 x - 3)}^{4}$ ?
The signs alternate, giving $16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81$ .
Treating $b$ as $(- 3)$ explains it: even powers of a negative number are positive and odd powers are negative.
Expand and simplify ${(3 x - 2)}^{5}$ .
Use the row 1, 5, 10, 10, 5, 1 with $a = 3 x$ and $b = - 2$ :
${(3 x - 2)}^{5} = 243 x^{5} - 810 x^{4} + 1080 x^{3} - 720 x^{2} + 240 x - 32$
Apply each power to the number as well as the letter, so that ${(3 x)}^{4} = 81 x^{4}$ .
How do you find a single term of an expansion without expanding all of it?
Work out which position the term you want sits in, then take just the matching Pascal coefficient and powers.
Every term has the form (Pascal coefficient) $\times a^{(\dots)} \times b^{(\dots)}$ , so only that one term needs calculating.
In the expansion of ${(a + b)}^{6}$ , which term contains $a^{4}$ ?
The one with $b^{2}$ , since the two powers must add up to 6.
Counting along from $a^{6}$ it is the third term, so its Pascal coefficient is the third entry of 1, 6, 15, 20, 15, 6, 1, which is 15.
Complete the $x^{3}$ term in the expansion of ${(4 x - 2)}^{5}$ :
$10 \times {(4 x)}^{______} \times {(- 2)}^{______}$
The completed term is:
$10 \times {(4 x)}^{3} \times {(- 2)}^{2} = 2560 x^{3}$
The two powers must add up to 5, and the $x^{3}$ has to come from ${(4 x)}^{3}$ , so the coefficient is $10 \times 64 \times 4 = 2560$ .
True or False?
In the expansion of ${(x + \frac{1}{x})}^{4}$ , the constant term is in the middle.
True.
The expansion is $x^{4} + 4 x^{2} + 6 + \frac{4}{x^{2}} + \frac{1}{x^{4}}$ , so the 6 sits third out of five.
Once the powers of $x$ cancel, the constant term no longer has to be at the end.
Why do powers of $x$ cancel when expanding ${(x + \frac{1}{x})}^{4}$ ?
Because both terms of the binomial involve $x$ , so each term of the expansion multiplies a power of $x$ by a power of $\frac{1}{x}$ .
For instance $4 x^{3} \times \frac{1}{x} = 4 x^{2}$ , using ${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$ on the fraction.
Given that $p > 0$ and the coefficient of $x^{4}$ in the expansion of ${(3 x - p)}^{6}$ is 59535, find $p$ .
The $x^{4}$ term is $15 {(3 x)}^{4} {(- p)}^{2}$ , so its coefficient is $15 \times 81 \times p^{2} = 1215 p^{2}$ .
Setting $1215 p^{2} = 59535$ gives $p^{2} = 49$ , and since $p > 0$ the answer is $p = 7$ .
Find the coefficient of $x$ in the expansion of ${(2 x + \frac{1}{x})}^{5}$ .
You need the term where ${(2 x)}^{3} {(\frac{1}{x})}^{2}$ leaves a single $x$ , since $8 x^{3}$ divided by $x^{2}$ is $8 x$ .
Its Pascal coefficient is the third entry of 1, 5, 10, 10, 5, 1, so the coefficient is $10 \times 8 = 80$ .

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