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Cambridge IGCSE0606

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Mathematics Additional 0606 Chapter Notes

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In this chapter

  1. 1The binomial theorem
  2. 2The general term and the term independent of x
  3. 3Products of expansions
  4. 4Arithmetic progressions
  5. 5Geometric progressions
  6. 6Convergence and the sum to infinity

1. The binomial theorem

To expand (a+b)ⁿ for a positive integer n, use the binomial theorem. The coefficients come from binomnr or from Pascal's triangle.

The powers of a go down by one each term and the powers of b go up by one. Work out every coefficient and write each term in its simplest form.

When a or b has a number or a power in it, put the whole bracket in. For (2x)³ write 8x³, not 2x³.

(a+b)ⁿ = aⁿ + dbinomn1aⁿ⁻¹b + dbinomn2aⁿ⁻²b² + ⋯ + bⁿ

dbinomnr = (n!)/(r! (n-r)!)

Examiner insight

“Some candidates may have improved if they had taken note of the instruction to give each term in its simplest form.” “Some candidates gave the first three terms in ascending powers of x.”

Worked example 13 marks

Find, in descending powers of x, the first three terms in the expansion of (x² + 1/2x)⁹. Give each term in its simplest form.

  1. 1

    first term: (x²)⁹ = x¹⁸ [1]

  2. 2

    second term: 9 × (x²)⁸ × 1/2x = 4.5x¹⁵ [1]

  3. 3

    third term: 36 × (x²)⁷ × 1/4x² = 9x¹² [1]

2. The general term and the term independent of x

The term in a^(n-r)b^(r) is the general term, for 0 leqslant r leqslant n. Use it to pick out one term without expanding the rest.

The term independent of x is the term with no x in it. It is the number term. Find the power of x in the general term, put it equal to zero and solve for r.

The value of r must be a whole number from 0 to n.

general term = dbinomnra^(n-r)b^(r)

independent of x: power of x = 0

Examiner insight

“Candidates should be aware that the term independent of x in an expansion is the number term.” “Few candidates stated that r had to be an integer, or equivalent, with completely correct working.”

Worked example 14 marks

The expansion of (x² + 1/2x)ⁿ, where n is a positive integer, has a term that is independent of x. Show that n is divisible by 3.

  1. 1

    general term: dbinomnr(x²)^(n-r)(1/2x)^(r) [1]

  2. 2

    power of x is 2(n-r) - r; independent of x means 2(n-r) = r [1]

  3. 3

    2n = 3r, so r = 2n/3 [1]

  4. 4

    r is an integer, so 2n is divisible by 3, so n is divisible by 3 [1]

3. Products of expansions

Some questions multiply two expansions. Expand each bracket first, to at least three terms.

Then list the pairs of terms, one from each bracket, whose powers of x add to zero. Multiply each pair and add the results.

One term is rarely enough. Look for two or three pairs.

Examiner insight

“Candidates had to consider three products using their answer to the first part and many did this.” “It is suggested that in questions of this type, candidates attempt to expand the binomial expressions to at least three terms even if they are unsure of how to proceed, as this will usually mean that they will obtain some marks.”

Worked example 16 marks

Find the term independent of x in the expansion of (2 - 3x)⁶(1 + 1/4x)².

  1. 1

    (2-3x)⁶ = 64 - 576x + 2160x² + ⋯, since 6 × 2⁵ × (-3) = -576 and 15 × 2⁴ × 9 = 2160 [2]

  2. 2

    (1+1/4x)² = 1 + 1/2x + 1/16x² [1]

  3. 3

    pairs with power zero: 64 × 1, -576x × 1/2x, 2160x² × 1/16x² [2]

  4. 4

    64 - 288 + 135 = -89 [1]

4. Arithmetic progressions

In an arithmetic progression each term is found by adding the same common difference d. The terms go up or down in equal steps.

The formulas are in the List of formulas. Use the nth term to find one term. Use the sum of the first n terms to add terms up. Do not swap them.

Two facts about the progression give two equations in a and d. Solve them together. For "the least number of terms" write an inequality, solve the quadratic and round up to a whole number.

un = a + (n-1)d

Sn = n/2{2a + (n-1)d}

Sn = n/2(a + l)

Examiner insight

“Some candidates may have improved if they had referred to the formula page at the start of the paper, as the expression for the nth term of an arithmetic progression was not always stated correctly.” “Weaker candidates sometimes used the formula for the sum of an arithmetic progression, rather than the nth term of an arithmetic progression.”

Worked example 18 marks

An arithmetic progression has first term a and common difference d.

The 6th term is 1.5 times the 3rd term.

The sum of the first ten terms is 255.

(a) Find the values of a and d.

(b) Using your values of a and d, find the least number of terms for the sum of this arithmetic progression to be greater than 40 000.

  1. 1

    6th term: a + 5d = 1.5(a + 2d), so a = 4d [1]

  2. 2

    sum of ten terms: 255 = 10/2(2a + 9d), so 2a + 9d = 51 [1]

  3. 3

    8d + 9d = 51, so d = 3 and a = 12 [2]

  4. 4

    n/2{24 + 3(n-1)} > 40 000 [1]

  5. 5

    3n² + 21n - 80 000 > 0 [1]

  6. 6

    solve the quadratic: n = (-21 + √(21² + 4 × 3 × 80 000))/6 = 159.8… [1]

  7. 7

    n = 160, the least whole number [1]

5. Geometric progressions

In a geometric progression each term is found by multiplying by the same common ratio r. Find r by dividing a term by the term before it.

A difference of terms is not a ratio. For 5, 6, 7.2, … the ratio is 6 ÷ 5 = 1.2, not 0.2.

The nth term has rⁿ⁻¹ in it. The sum of the first n terms is a fraction with 1 - rⁿ on top. Keep every answer exact unless the question asks for a decimal.

un = arⁿ⁻¹

Sn = (a(1 - rⁿ))/(1 - r)

Examiner insight

“Some confused the geometric progression with an arithmetic progression and found a common difference and then used the geometric progression formula.” “Candidates who gave non-exact decimal answers such as 0.471 needed to understand that this was not acceptable as it did not define the geometric progression accurately.”

Worked example 14 marks

A geometric progression has common ratio r and first term a, where a ≠ 0. The 4th term is -8 times the 7th term.

(i) Find the value of r.

(ii) The sum of the first 5 terms of this progression is 99/32. Find the value of a.

  1. 1

    4th term is -8 times the 7th term: ar³ = -8ar⁶ [1]

  2. 2

    divide by ar³: r³ = -1/8, so r = -1/2 [1]

  3. 3

    S₅ = (a(1 - (-1/2)⁵))/(1 - (-1/2)) = 99/32 [1]

  4. 4

    11a/16 = 99/32, so a = 9/2 [1]

6. Convergence and the sum to infinity

A geometric progression converges when |r| < 1, that is -1 < r < 1. Only then does it have a sum to infinity.

If |r| geqslant 1 it does not converge, so there is no sum to infinity.

To explain, state the common ratio, then state the condition. Write |r| < 1 and not r < 1. Check every value of r you find. Then use the formula.

S_∞ = a/(1 - r), |r| < 1

Examiner insight

“A few candidates stated that r < 1 which was not accepted.” “In weaker responses, candidates found a value for the sum to infinity and determined that, as they had found it, it existed.”

Worked example 18 marks

It is given that a geometric progression has the following terms.

(a) Given that x > 0, find the common ratio and first term of the progression.

(b) The sum to infinity of this progression is k/3 where k is an integer. Find the value of k.

termvalue
2nd term11x - 1
3rd termx + 5
4th termx - 1
  1. 1

    equal ratios: (x+5)/(11x-1) = (x-1)/(x+5) [1]

  2. 2

    (x+5)² = (x-1)(11x-1), so x² + 10x + 25 = 11x² - 12x + 1 [1]

  3. 3

    10x² - 22x - 24 = 0 [1]

  4. 4

    divide by 2 and factorise: (5x+4)(x-3) = 0 [1]

  5. 5

    x > 0, so x = 3 only [1]

  6. 6

    r = (x+5)/(11x-1) = 8/32 = 1/4, and 2nd term = ar = 32, so a = 128 [1]

  7. 7

    |r| < 1, so S_∞ = 128/(1 - 1/4) [1]

  8. 8

    S_∞ = 512/3, so k = 512 [1]

Recap

  • Binomial theorem: powers of a go down and powers of b go up, with coefficients binomnr.
  • General term: binomnra^(n-r)b^(r), with r a whole number from 0 to n.
  • Term independent of x = the number term: set the power of x to zero.
  • In a product of expansions, add up every pair of terms whose powers of x cancel.
  • Arithmetic: add d. un = a + (n-1)d and Sn = n/2{2a+(n-1)d}.
  • Geometric: multiply by r. un = arⁿ⁻¹ and Sn = (a(1-rⁿ))/(1-r).
  • Sum to infinity: S_∞ = a/(1-r) only if |r| < 1. State the condition.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State whether the sequence 3, 6, 12, 24, … is an arithmetic or a geometric progression. Give the common difference or the common ratio.2 marks
  2. Do not use a calculator. Expand (1 + 2x)⁴ in ascending powers of x.3 marks
  3. An arithmetic progression has first term 7 and common difference 4. (a) Find the 20th term. (b) Find the sum of the first 20 terms.4 marks
  4. Find the coefficient of x³ in the expansion of (2 - x)⁵.3 marks
  5. Do not use a calculator. A geometric progression has first term 5 and common ratio 3. Find the 6th term.2 marks
  6. Do not use a calculator. The 3rd term of an arithmetic progression is 11 and the 8th term is 26. (a) Find the first term and the common difference. (b) Find the sum of the first 12 terms.5 marks
  7. Do not use a calculator. A geometric progression has first term 18 and second term -6. Explain why it has a sum to infinity and find its value.4 marks
  8. A ball is dropped from a height of 2 m. After each bounce it rises to 0.6 of the height it has just fallen from. Assume that it goes on bouncing for ever. Find the total distance the ball travels.5 marks
  9. Do not use a calculator. The term independent of x in the expansion of (1 + kx³)(x - 2/x²)⁶ is -20, where k is a constant. Find the value of k.6 marks
  10. Do not use a calculator. A pointer starts at 0 on a number line. A move in the positive direction is positive and a move in the negative direction is negative. The first move is 20 units. Each later move is r times the move before it. The sum of the 2nd move and the 3rd move is 15 units. (a) Find the two possible values of r. (b) The pointer gets closer and closer to one fixed position. Explain which value of r this needs, and find that position. (c) For this value of r, find the least number of moves after which the pointer is less than 1 unit from that position.9 marks

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