Chapter 7 of 7

All 7 chapters
  1. 1Domain, range and inverses, readunderstood
  2. 2The modulus function, readunderstood
  3. 3Dividing polynomials, readunderstood
  4. 4The factor theorem, readunderstood
  5. 5Partial fractions, readunderstood
  6. 6Pascal's triangle and choosing, readunderstood
  7. 7The binomial expansion, readunderstood

M5.1.7 Chapter 7

The binomial expansion

Can I expand (1 + x)⁻¹ the way I expand (1 + x)²?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Expand (2 + x)⁵ with the binomial theorem
  • Find one term, like the x³ term, without the full expansion
  • Expand (1 + x)⁻¹ for small x and say when it is valid

Chapter 6 showed that the coefficient of a^(n − r) b^r in (a + b)ⁿ is nCr. (The sign ^ means "to the power". The book font has no raised r.) Written out, this is the binomial theorem.

(a + b)ⁿ = aⁿ + nC1 aⁿ⁻¹ b + nC2 aⁿ⁻² b² + nC3 aⁿ⁻³ b³ + … + bⁿ. It holds for every positive whole number n. The term with b to the power r is nCr a^(n − r) b^r.

Take a = 2 and b = x. Each term is nCr × 2 to the power (n − r) × x to the power r. The powers of 2 must be worked out. They are the usual place for a slip.

Pascal's triangle with six rows, from row 0 to row 5. Every entry in row 5 is ringed: 1, 5, 10, 10, 5 and 1.11112113311464115101051
Row 5 holds the numbers 5C0 to 5C5. They are the coefficients before the powers of 2 are put in.
A table with the heads r, 5Cr, 2 power and term, where 2 power is the power of 2. Rows: r = 0, 5Cr = 1, power of 2 = 32, term 32; r = 1, 5, 16, 80x; r = 2, 10, 8, 80x²; r = 3, 10, 4, 40x³; r = 4, 5, 2, 10x⁴; r = 5, 1, 1, x⁵.r5Cr2 powerterm013232151680x210880x²310440x³45210x⁴511x⁵
The five columns of work for (2 + x)⁵. The term is 5Cr times the power of 2 times x to the power r.

Worked example

Expand (2 + x)⁵ in ascending powers of x. (Ascending means the lowest power of x first.)

  1. Here a = 2, b = x and n = 5. The coefficients 5C0 to 5C5 are 1, 5, 10, 10, 5, 1 (row 5 of Pascal's triangle).
  2. The power of 2 falls from 5 to 0: 2⁵ = 32, 2⁴ = 16, 2³ = 8, 2² = 4, 2¹ = 2, 2⁰ = 1.
  3. Multiply each coefficient by its power of 2: 1 × 32 = 32, 5 × 16 = 80, 10 × 8 = 80, 10 × 4 = 40, 5 × 2 = 10, 1 × 1 = 1.
  4. Attach x to the power r: 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵.
  5. Check with x = 1. The left side is 3⁵ = 243. The right side is 32 + 80 + 80 + 40 + 10 + 1 = 243.

Answer: 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵

Often we want only one term. The general term, the one with b to the power r, is the (r + 1)th term. It is written T(r + 1) = nCr a^(n − r) b^r. Put in the values of n, r, a and b. There is no need to write the whole expansion.

Take care when b has a number in it. The whole of b is raised to the power r. So (2x)³ = 2³ × x³ = 8x³, and (−3x)² = 9x².

A table with the heads part and value. Rows: 5C3 is 10; 3² is 9; (2x)³ is 8x³; the product is 720x³.partvalue5C3103²9(2x)³8x³product720x³
The three parts of the x³ term of (3 + 2x)⁵, and their product.

Worked example

Find the coefficient of x³ in the expansion of (3 + 2x)⁵.

  1. Here a = 3, b = 2x and n = 5. The term with x³ has r = 3.
  2. The general term is 5C3 × 3² × (2x)³, because the power of a is n − r = 2.
  3. Work out the parts: 5C3 = 10, 3² = 9, and (2x)³ = 8x³.
  4. Multiply: 10 × 9 × 8 = 720.
  5. So the x³ term is 720x³, and the coefficient is 720.

Answer: The coefficient of x³ is 720.

Worked example

Find the coefficient of x² in the expansion of (1 + 2x)(1 − x)⁶.

  1. First expand (1 − x)⁶ as far as x². Here a = 1 and b = −x, so the terms are 6C0, 6C1(−x), 6C2(−x)².
  2. 6C0 = 1. 6C1 × (−x) = −6x. 6C2 × (−x)² = 15x², because (−x)² = x².
  3. So (1 − x)⁶ = 1 − 6x + 15x² + …
  4. Now multiply by (1 + 2x). The x² terms come from two products: 1 × 15x², and 2x × (−6x).
  5. These are 15x² and −12x². Their sum is 15 − 12 = 3.
  6. So the coefficient of x² is 3.

Answer: The coefficient of x² is 3.

Now the power can be any rational number, not only a positive whole number. A rational number is a fraction of two whole numbers, with the bottom not zero. This includes −1, −2 and 1/2.

The coefficients nC1, nC2, nC3 are n, n(n − 1)/2!, n(n − 1)(n − 2)/3!. These products make sense for every n, not only whole numbers. This gives the general result.

For any rational n, (1 + x)ⁿ = 1 + nx + [n(n − 1)/2!] x² + [n(n − 1)(n − 2)/3!] x³ + …, and it is valid for |x| < 1, which means −1 < x < 1.

There are two big differences from a positive whole number n. First, the series never stops: for n = −1 no factor n − r is ever 0. Second, it works only for |x| < 1. For a positive whole n, the factor (n − n) = 0 stops the series after n + 1 terms, and then it is true for every x.

Take n = −1. The coefficients are 1, then n = −1, then (−1)(−2)/2! = 1, then (−1)(−2)(−3)/3! = −1. So (1 + x)⁻¹ = 1 − x + x² − x³ + …

Why |x| < 1? The terms 1, −x, x², −x³ are a geometric series with first term 1 and common ratio −x. A geometric series adds up to 1/(1 − (−x)) = 1/(1 + x) only when the ratio is between −1 and 1, that is, |x| < 1.

A table with the heads x, exact and series. Exact is 1/(1 + x) and series is 1 − x + x² − x³. Rows: x = 0.1, 0.9091, 0.9090; x = 0.5, 0.6667, 0.6250; x = 2, 0.3333, −5.0000.xexactseries0.10.90910.90900.50.66670.625020.3333−5.0000
Four terms of the series for 1/(1 + x). The sum is close when x is small, rough at x = 0.5, and far wrong at x = 2.
A graph with x from −1 to 1 and y from 0 to 4. A solid curve named f is the graph of 1/(1 + x). It climbs very steeply as x gets near −1. A dashed curve named g is the graph of 1 − x + x² − x³. The two curves lie very close together near x = 0 and move apart toward the left.fg−1−0.50.512340xy
f(x) = 1/(1 + x) and g(x) = 1 − x + x² − x³. Near x = 0 the two agree. Towards x = −1 the series falls far short.

Worked example

Expand (1 + x)⁻¹ in ascending powers of x as far as x³. State the values of x for which the expansion is valid, and test it at x = 0.1.

  1. Use (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + … with n = −1.
  2. The x term is nx = −x.
  3. The x² coefficient is (−1)(−2)/2 = 1.
  4. The x³ coefficient is (−1)(−2)(−3)/6 = −1.
  5. So (1 + x)⁻¹ = 1 − x + x² − x³ + …, valid for |x| < 1.
  6. Test at x = 0.1: the left side is 1/1.1 = 0.90909… The right side is 1 − 0.1 + 0.01 − 0.001 = 0.909, close to the left.

Answer: 1 − x + x² − x³ + …, valid for |x| < 1.

To expand (a + bx)ⁿ for a negative or fractional n, first take out the number a: (a + bx)ⁿ = aⁿ(1 + (b/a)x)ⁿ. Then expand (1 + (b/a)x)ⁿ. This needs a > 0 when n is not a whole number, so that aⁿ makes sense. The expansion is valid when |(b/a)x| < 1, which means |x| < |a/b|.

For (1 + 3x)⁻², put 3x in place of x. The series is valid for |3x| < 1, which means |x| < 1/3. The limit is 1/3, not 1. Always divide: the limit for |bx| < 1 is |x| < 1/|b|.

Worked example

Expand (1 + 3x)⁻² in ascending powers of x up to x³. State the range of x for which it is valid. Hence find the coefficient of x² in the expansion of (3 − x)(1 + 3x)⁻².

  1. Put 3x in place of x and n = −2 in (1 + x)ⁿ. The coefficients are n = −2, then (−2)(−3)/2 = 3, then (−2)(−3)(−4)/6 = −4.
  2. The terms are 1, then −2 × 3x = −6x, then 3 × (3x)² = 3 × 9x² = 27x², then −4 × (3x)³ = −4 × 27x³ = −108x³.
  3. So (1 + 3x)⁻² = 1 − 6x + 27x² − 108x³ + …
  4. The expansion is valid for |3x| < 1, which means |x| < 1/3.
  5. For the product (3 − x)(1 + 3x)⁻², the x² terms are 3 × 27x² and (−x) × (−6x).
  6. These are 81x² and 6x². So the coefficient of x² is 81 + 6 = 87.

Answer: 1 − 6x + 27x² − 108x³, valid for |x| < 1/3, and the coefficient is 87.

A case worth knowing is n = 1/2, the square root. Its coefficients are 1, then n = 1/2, then (1/2)(−1/2)/2! = −1/8, then (1/2)(−1/2)(−3/2)/3! = 1/16. So √(1 + x) = 1 + x/2 − x²/8 + x³/16 − …, valid for |x| < 1. For example √1.02 is about 1 + 0.01 − 0.00005 = 1.00995.

Worked example

Expand (1 − 2x)⁻¹ in ascending powers of x up to x³. State when it is valid.

  1. We know (1 + y)⁻¹ = 1 − y + y² − y³ + …, valid for |y| < 1.
  2. Put y = −2x. Then −y = 2x, y² = 4x² and −y³ = 8x³.
  3. So (1 − 2x)⁻¹ = 1 + 2x + 4x² + 8x³ + …
  4. The condition |y| < 1 becomes |−2x| < 1, which is |2x| < 1. So |x| < 1/2.
  5. Check with x = 0.1. The series gives 1 + 0.2 + 0.04 + 0.008 = 1.248. The left side is 1/0.8 = 1.25, which is close.

Answer: 1 + 2x + 4x² + 8x³ + …, valid for |x| < 1/2.

Worked example

Expand √(4 − x) in ascending powers of x up to x². State when it is valid. Use x = 0.04 to estimate √3.96.

  1. Take out the 4: √(4 − x) = √4 × √(1 − x/4) = 2(1 − x/4)^(1/2).
  2. Use (1 + y)ⁿ with n = 1/2 and y = −x/4: 1 + ny + n(n − 1)/2! y² + …
  3. The y term is (1/2)(−x/4) = −x/8.
  4. The y² coefficient is (1/2)(−1/2)/2 = −1/8. And y² = x²/16, so the term is −x²/128.
  5. So (1 − x/4)^(1/2) = 1 − x/8 − x²/128 + … Multiply by 2: √(4 − x) = 2 − x/4 − x²/64 + …
  6. It is valid for |−x/4| < 1, which means |x| < 4.
  7. Put x = 0.04. Then √3.96 is about 2 − 0.04/4 − 0.0016/64 = 2 − 0.01 − 0.000025 = 1.989975.

Answer: 2 − 1/4x − 1/64x², valid for |x| < 4, and √3.96 is about 1.989975.

Another favourite is the term independent of x, the one with no x in it. In a general term, collect the power of x, set it equal to 0, and solve for r. If r is not a whole number from 0 to n, there is no such term.

Worked example

Find the term independent of x in the expansion of (x + 2/x)⁶.

  1. Here a = x, b = 2/x and n = 6. The general term is T(r + 1) = 6Cr × x^(6 − r) × (2/x)^r.
  2. Write (2/x)^r = 2^r × x^(−r). So the term is 6Cr × 2^r × x^(6 − r − r) = 6Cr × 2^r × x^(6 − 2r).
  3. The term has no x when the power is 0: 6 − 2r = 0, so r = 3.
  4. Put r = 3: 6C3 = 20 and 2³ = 8.
  5. So the term is 20 × 8 = 160, and that is the term independent of x.

Answer: The term independent of x is 160.

Try it

Test an expansion with numbers

You need: paper, a pencil, a calculator if you have one.

  1. Take the result (2 + x)⁵ = 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵.
  2. Put x = 1, then x = 2, then x = −1 into the left side. Write each answer.
  3. Put the same values into the right side and add the terms.
  4. Compare. A wrong coefficient will almost always change one of the answers.

What you will see: The two sides agree for all three values. Putting in x = 1 is the quickest check, because it just adds the coefficients.

If you cannot try it, this is what happens.
A table with the heads x, left side and right side. Rows: x = 1, left 243, right 243; x = 2, left 1 024, right 1 024; x = −1, left 1, right 1.xleftright124324321 0241 024−111

A wrong idea many people have

(2 + x)⁵ = 2 + 5x + 10x² + 10x³ + 5x⁴ + x⁵, using the row of the triangle only.

Why it feels right: The coefficients of (1 + x)⁵ are 1, 5, 10, 10, 5, 1, and the 2 seems to be just one more number.

The test that breaks it: Put x = 1. The left side is 3⁵ = 243. This right side adds to 2 + 5 + 10 + 10 + 5 + 1 = 33, which is not 243.

What is true: Every term needs the power of 2: 2⁵, 2⁴, 2³, and so on. The coefficient of x is 5 × 2⁴ = 80, not 5.

A wrong idea many people have

In (3 + 2x)⁵ the x³ term uses 2x³, because the 2 is just a number in front.

Why it feels right: The 2 is next to the x, so only x seems to get the power 3.

The test that breaks it: (2x)³ = 2x × 2x × 2x = 8x³. The whole of 2x is cubed, so 2 is cubed too.

What is true: Raise the whole of b to the power r: (2x)³ = 8x³. Then the coefficient is 10 × 3² × 8 = 720, not 10 × 9 × 2 = 180.

A wrong idea many people have

The series for (1 + x)⁻¹ gives the right answer for every x.

Why it feels right: For a positive whole n, the expansion is true for every x, so it seems the same for all n.

The test that breaks it: Put x = 2. The series gives 1 − 2 + 4 − 8 + 16 − …, which grows without limit. But 1/(1 + 2) = 1/3.

What is true: For n not a positive whole number, the series is valid only for |x| < 1, and for (1 + bx)ⁿ only for |bx| < 1. State the range every time.

A wrong idea many people have

The expansion of (1 + x)⁻¹ stops after a few terms, as (1 + x)³ does.

Why it feels right: Any power of a bracket seems to have a last term, as bⁿ does.

The test that breaks it: The coefficients for n = −1 are 1, −1, 1, −1, … They are never 0. For n = 3 the factor (n − 3) is 0, which stops the series.

What is true: Only a positive whole number n gives a finite expansion. Any other n gives a series that goes on for ever, which is why its range of validity must be stated.

Key words

binomial theorem
the formula for (a + b)ⁿ as a sum of terms nCr a^(n − r) b^r, for a positive whole number n.
general term
the term with b to the power r in the expansion, which lets you find one term without the rest.
ascending powers
written with the lowest power of x first.
valid
giving the right value. The series for (1 + x)ⁿ is valid only for |x| < 1 unless n is a positive whole number.

Back to your guess

The answer is C: Yes, but the series never ends, and it works only when x is small enough.

It can be expanded: 1/(1 + x) = 1 − x + x² − x³ + …. The terms go on for ever, so the series never ends. And it gives the right value only when x lies between −1 and 1. For x = 2 the terms 1, −2, 4, −8 grow without limit, but 1/(1 + 2) is just a third.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. What is the coefficient of x² in the expansion of (1 + 3x)⁴? The picture shows Pascal's triangle to row 4.

    Pascal's triangle with five rows, from row 0 to row 4. Row 4 is 1, 4, 6, 4, 1.111121133114641
    Tap your answer.

    1. Not this one. You multiplied 6 by 3 once. The x² term has (3x)², and 3² = 9, so the whole power applies to the 3. Go to M5.1.7. Try another.

    2. Not this one. You used the coefficient 4 from row 4, place 1. The x² term uses place 2, which is 6. Go to M5.1.7. Try another.

    3. Not this one. That is 3⁴, the whole power. The x² term has (3x)² = 9x², and it is multiplied by 4C2. Go to M5.1.7. Try another.

    4. Yes, that is right. The x² term is 4C2 × 1² × (3x)² = 6 × 9x² = 54x². So the coefficient is 54.

    See the answer

    54

    The x² term is 4C2 × 1² × (3x)² = 6 × 9x² = 54x². So the coefficient is 54.

    • If you chose A: You multiplied 6 by 3 once. The x² term has (3x)², and 3² = 9, so the whole power applies to the 3. Go to M5.1.7.
    • If you chose B: You used the coefficient 4 from row 4, place 1. The x² term uses place 2, which is 6. Go to M5.1.7.
    • If you chose C: That is 3⁴, the whole power. The x² term has (3x)² = 9x², and it is multiplied by 4C2. Go to M5.1.7.

  2. What are the first two terms of the expansion of (3 + x)⁴ in ascending powers of x?

    Tap your answer.

    1. Yes, that is right. The first term is 3⁴ = 81. The second is 4C1 × 3³ × x = 4 × 27 × x = 108x.

    2. Not this one. You used 3 instead of 3³ in the x term. The x term is 4C1 × 3³ × x = 4 × 27x = 108x. Go to M5.1.7. Try another.

    3. Not this one. That drops the powers of 3 and the coefficients. Each term has a^(n − r) with a = 3, so the first term is 3⁴ = 81. Go to M5.1.7. Try another.

    4. Not this one. Check the first term: it is a⁴ = 3⁴ = 81, with coefficient 1, not 4 × 3. Go to M5.1.7. Try another.

    See the answer

    81 + 108x

    The first term is 3⁴ = 81. The second is 4C1 × 3³ × x = 4 × 27 × x = 108x.

    • If you chose B: You used 3 instead of 3³ in the x term. The x term is 4C1 × 3³ × x = 4 × 27x = 108x. Go to M5.1.7.
    • If you chose C: That drops the powers of 3 and the coefficients. Each term has a^(n − r) with a = 3, so the first term is 3⁴ = 81. Go to M5.1.7.
    • If you chose D: Check the first term: it is a⁴ = 3⁴ = 81, with coefficient 1, not 4 × 3. Go to M5.1.7.

  3. What is the coefficient of x² in the expansion of (1 + x)⁻²?

    Tap your answer.

    1. Not this one. Check the sign: n(n − 1)/2 = (−2)(−3)/2 = 6/2. Two negative numbers multiply to a positive. Go to M5.1.7. Try another.

    2. Not this one. You forgot to divide by 2!. The coefficient is n(n − 1)/2!, which is 6/2. Go to M5.1.7. Try another.

    3. Yes, that is right. The x² coefficient is n(n − 1)/2! with n = −2. That is (−2)(−3)/2 = 6/2 = 3.

    4. Not this one. That is the coefficient of x² in (1 + x)⁻¹. For n = −2 the coefficient is n(n − 1)/2! = (−2)(−3)/2. Go to M5.1.7. Try another.

    See the answer

    3

    The x² coefficient is n(n − 1)/2! with n = −2. That is (−2)(−3)/2 = 6/2 = 3.

    • If you chose A: Check the sign: n(n − 1)/2 = (−2)(−3)/2 = 6/2. Two negative numbers multiply to a positive. Go to M5.1.7.
    • If you chose B: You forgot to divide by 2!. The coefficient is n(n − 1)/2!, which is 6/2. Go to M5.1.7.
    • If you chose D: That is the coefficient of x² in (1 + x)⁻¹. For n = −2 the coefficient is n(n − 1)/2! = (−2)(−3)/2. Go to M5.1.7.

  4. For which values of x is the expansion of (1 + 5x)⁻¹ valid?

    Tap your answer.

    1. Not this one. You multiplied where you should divide. The condition is |5x| < 1, and dividing both sides by 5 gives |x| < 1/5. Go to M5.1.7. Try another.

    2. Not this one. That is the condition for (1 + x)⁻¹. Here 5x takes the place of x, so |5x| < 1. Go to M5.1.7. Try another.

    3. Not this one. Only a positive whole number power gives an expansion that is true for every x. A power of −1 gives a series that goes on for ever. Go to M5.1.7. Try another.

    4. Yes, that is right. The expansion of (1 + y)⁻¹ is valid for |y| < 1. Here y = 5x, so |5x| < 1. Divide by 5: |x| < 1/5.

    See the answer

    |x| < 1/5

    The expansion of (1 + y)⁻¹ is valid for |y| < 1. Here y = 5x, so |5x| < 1. Divide by 5: |x| < 1/5.

    • If you chose A: You multiplied where you should divide. The condition is |5x| < 1, and dividing both sides by 5 gives |x| < 1/5. Go to M5.1.7.
    • If you chose B: That is the condition for (1 + x)⁻¹. Here 5x takes the place of x, so |5x| < 1. Go to M5.1.7.
    • If you chose C: Only a positive whole number power gives an expansion that is true for every x. A power of −1 gives a series that goes on for ever. Go to M5.1.7.

  5. Find the coefficient of x² in the expansion of (2 − 3x)⁴.

    See the answer

    216

    The x² term is 4C2 × 2² × (−3x)² = 6 × 4 × 9x² = 216x². Here (−3x)² = 9x², so the minus sign goes.

  6. The expansion of (2 + 5x)⁻¹ in ascending powers of x is valid for |x| < c. Find c.

    See the answer

    2/5

    Take out the 2: (2 + 5x)⁻¹ = 2⁻¹(1 + 5x/2)⁻¹. The expansion of (1 + y)⁻¹ is valid for |y| < 1. Here y = 5x/2, so |5x/2| < 1. Multiply by 2 and divide by 5: |x| < 2/5, so c = 2/5.

  7. (a) Write (4 + 5x)/((1 − x)(1 + 2x)) as A/(1 − x) + B/(1 + 2x). (b) Hence expand it in ascending powers of x up to x². (c) State the values of x for which the expansion is valid. Give A and B, then the expansion, then the limit of |x|.

    See the answer

    A = 3 and B = 1. The expansion is 4 + x + 7x², valid for |x| < 1/2.

    (a) 4 + 5x = A(1 + 2x) + B(1 − x). Put x = 1: 9 = 3A, so A = 3. Put x = −1/2: 4 − 5/2 = 3/2 = B × 3/2, so B = 1. (b) 3/(1 − x) = 3 + 3x + 3x² + … And 1/(1 + 2x) = 1 − 2x + 4x² − … Add: 4 + x + 7x² + … (c) The first part needs |x| < 1. The second needs |2x| < 1. Both hold only when |x| < 1/2.

  8. From an earlier chapter

    A coach chooses 3 players from a squad of 10 to take a penalty. In how many ways can she choose them?

    Tap your answer.

    1. Not this one. That is 10 × 9 × 8, which counts the players in order. The order does not matter, so divide by 3! = 6. Go to M5.1.6. Try another.

    2. Yes, that is right. The order does not matter, so the number of ways is 10C3 = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120.

    3. Not this one. You multiplied 10 by 3. Use 10C3 = (10 × 9 × 8)/(3 × 2 × 1). Go to M5.1.6. Try another.

    4. Not this one. That is 10 × 10 × 10, which would allow the same player to be chosen three times. Go to M5.1.6. Try another.

    See the answer

    120

    The order does not matter, so the number of ways is 10C3 = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120.

    • If you chose A: That is 10 × 9 × 8, which counts the players in order. The order does not matter, so divide by 3! = 6. Go to M5.1.6.
    • If you chose C: You multiplied 10 by 3. Use 10C3 = (10 × 9 × 8)/(3 × 2 × 1). Go to M5.1.6.
    • If you chose D: That is 10 × 10 × 10, which would allow the same player to be chosen three times. Go to M5.1.6.

  9. What is the coefficient of x² in the expansion of (1 − 2x)^(1/2)?

    Tap your answer.

    1. Not this one. That is the coefficient of y² in (1 + y)^(1/2). Here y = −2x, so y² = 4x² and the coefficient of x² is 4 times as big. Go to M5.1.7. Try another.

    2. Not this one. Check the sign: n(n − 1)/2! = (1/2)(−1/2)/2 is negative. The square of −2x is positive, but the coefficient itself is negative. Go to M5.1.7. Try another.

    3. Not this one. That is the coefficient of x, which is n × (−2) = −1, not the x² coefficient. The x² term uses n(n − 1)/2! and (−2x)². Go to M5.1.7. Try another.

    4. Yes, that is right. Put n = 1/2 and y = −2x in (1 + y)ⁿ. The y² coefficient is (1/2)(−1/2)/2! = −1/8. And y² = (−2x)² = 4x². So the x² term is −1/8 × 4x² = −1/2 x², and the coefficient is −1/2.

    See the answer

    −1/2

    Put n = 1/2 and y = −2x in (1 + y)ⁿ. The y² coefficient is (1/2)(−1/2)/2! = −1/8. And y² = (−2x)² = 4x². So the x² term is −1/8 × 4x² = −1/2 x², and the coefficient is −1/2.

    • If you chose A: That is the coefficient of y² in (1 + y)^(1/2). Here y = −2x, so y² = 4x² and the coefficient of x² is 4 times as big. Go to M5.1.7.
    • If you chose B: Check the sign: n(n − 1)/2! = (1/2)(−1/2)/2 is negative. The square of −2x is positive, but the coefficient itself is negative. Go to M5.1.7.
    • If you chose C: That is the coefficient of x, which is n × (−2) = −1, not the x² coefficient. The x² term uses n(n − 1)/2! and (−2x)². Go to M5.1.7.

  10. The coefficient of x² in the expansion of (1 + ax)⁶ is 60, and a is positive. Find a.

    See the answer

    2

    The x² term is 6C2 × (ax)² = 15a²x². So 15a² = 60, which gives a² = 4. Since a is positive, a = 2.

  11. Find the term independent of x in the expansion of (2x − 1/x)⁴.

    Tap your answer.

    1. Yes, that is right. The general term is 4Cr × (2x)^(4 − r) × (−1/x)^r = 4Cr × 2^(4 − r) × (−1)^r × x^(4 − 2r). The power of x is 0 when r = 2. So the term is 4C2 × 2² × (−1)² = 6 × 4 × 1 = 24.

    2. Not this one. Check the sign: the term has (−1/x)², and a minus sign squared is plus. So the term is positive. Go to M5.1.7. Try another.

    3. Not this one. You used 4C2 = 6 and forgot the powers of 2x and −1/x. The term is 4C2 × (2x)² × (−1/x)². Go to M5.1.7. Try another.

    4. Not this one. You used (2x)⁴, which is the r = 0 term. For no x we need r = 2, so the power of 2x is 2, and (2x)² = 4x². Go to M5.1.7. Try another.

    See the answer

    24

    The general term is 4Cr × (2x)^(4 − r) × (−1/x)^r = 4Cr × 2^(4 − r) × (−1)^r × x^(4 − 2r). The power of x is 0 when r = 2. So the term is 4C2 × 2² × (−1)² = 6 × 4 × 1 = 24.

    • If you chose B: Check the sign: the term has (−1/x)², and a minus sign squared is plus. So the term is positive. Go to M5.1.7.
    • If you chose C: You used 4C2 = 6 and forgot the powers of 2x and −1/x. The term is 4C2 × (2x)² × (−1/x)². Go to M5.1.7.
    • If you chose D: You used (2x)⁴, which is the r = 0 term. For no x we need r = 2, so the power of 2x is 2, and (2x)² = 4x². Go to M5.1.7.

  12. Use the first three terms of the expansion of (1 + x)^(1/2) to estimate √1.04. Give your answer to 4 decimal places.

    See the answer

    1.0198

    Put x = 0.04 in 1 + x/2 − x²/8. That is 1 + 0.02 − 0.0002 = 1.0198. (The true value is 1.0198039…, so the estimate is good to 4 places.)

Keep it

Copy this into your exercise book.

(a + b)ⁿ: term nCr a^(n − r) b^r. Raise all of b to r. For any other n: (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + …, endless, valid for |x| < 1.

Draw: A line of expansion terms with nCr above each, and the number line from −1 to 1 marked as the valid range.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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