A1Arena Open the app

Sequences, series and binomial expansion

Further Mathematics · WAEC and JAMB · SS2 and SS3

Arithmetic and geometric progressions appear in almost every paper, and the binomial theorem is a reliable source of one full question. Both reward candidates who write down the correct general term first and substitute afterwards.

What you need to know

  • In an arithmetic progression each term is obtained by adding a fixed common difference d. The nth term is T_n = a + (n - 1)d, where a is the first term, so the coefficient of d is always one less than the term number.
  • The sum of the first n terms of an AP is S_n = n/2 [2a + (n - 1)d], or equivalently S_n = n/2 (a + L) when you know the last term L. Use the second form whenever the final term is given, because it is faster.
  • When a question gives you two terms of an AP, write each as an equation in a and d, then subtract to eliminate a. For example T_5 = 11 and T_9 = 23 give a + 4d = 11 and a + 8d = 23, so 4d = 12 and d = 3.
  • In a geometric progression each term is obtained by multiplying by a fixed common ratio r. The nth term is T_n = a r^(n-1), and dividing any term by the one before it gives r directly.
  • The sum of the first n terms of a GP is S_n = a(r^n - 1)/(r - 1) when r is greater than 1, and S_n = a(1 - r^n)/(1 - r) when r is less than 1. The two forms are identical, so use whichever keeps your denominator positive.
  • A GP has a sum to infinity only when the absolute value of r is less than 1, and then S = a/(1 - r). With a = 3 and r = 1/2 the sum to infinity is 3/(1 - 1/2) = 6. If r is 1 or more, state that the sum to infinity does not exist.
  • A recurring decimal is a geometric series. 0.4444... is 0.4 + 0.04 + 0.004 + ... with a = 0.4 and r = 0.1, so its value is 0.4/0.9 = 4/9.
  • The arithmetic mean of a and b is (a + b)/2 and the geometric mean is sqrt(ab). When a question says three numbers are in GP, use the fact that the middle term squared equals the product of the outer two.
  • The binomial theorem states (a + b)^n = sum of nCr a^(n-r) b^r for r from 0 to n, where nCr = n!/(r!(n - r)!). The powers of a fall from n to 0 while the powers of b rise from 0 to n, and the two powers always add up to n.
  • The general term, often written T_(r+1) = nCr a^(n-r) b^r, is the key to every find the term question. Write it out with the given a, b and n, simplify the power of x, then set that power equal to what the question wants.
  • For a term independent of x, set the final power of x equal to zero and solve for r. For the coefficient of x^k, set the power equal to k. Only after finding r do you compute the numerical coefficient.
  • Pascal's triangle gives the coefficients quickly for small n, and the row for n = 5 is 1, 5, 10, 10, 5, 1. For n above about 6 use nCr instead, since counting rows wastes time and invites error.
  • For small values of x, (1 + x)^n is approximately 1 + nx + n(n - 1)x^2/2. This is how you estimate quantities such as (1.02)^8 without a calculator, by writing it as (1 + 0.02)^8.

Key terms

Sequence
An ordered list of numbers generated by a rule, where each number is called a term.
Series
The sum of the terms of a sequence.
Common difference
The fixed amount added to each term of an arithmetic progression to get the next term.
Common ratio
The fixed amount each term of a geometric progression is multiplied by to get the next term.
Convergent series
A series whose partial sums approach a finite limit, which for a GP happens exactly when the absolute value of r is less than 1.
Binomial coefficient
The number nCr = n!/(r!(n - r)!), giving the coefficient of the term containing b^r in the expansion of (a + b)^n.

Formulae

  • AP nth term: T_n = a + (n - 1)d
  • AP sum: S_n = n/2 [2a + (n - 1)d]
  • AP sum with last term: S_n = n/2 (a + L)
  • GP nth term: T_n = a r^(n-1)
  • GP sum: S_n = a(r^n - 1)/(r - 1) for r > 1
  • GP sum: S_n = a(1 - r^n)/(1 - r) for r < 1
  • GP sum to infinity: S = a/(1 - r), valid only when |r| < 1
  • Arithmetic mean of a and b = (a + b)/2
  • Geometric mean of a and b = sqrt(ab)
  • nCr = n!/(r!(n - r)!)
  • (a + b)^n = sum over r of nCr a^(n-r) b^r
  • General term: T_(r+1) = nCr a^(n-r) b^r
  • (1 + x)^n is approximately 1 + nx + n(n-1)x^2/2 for small x

Worked examples

The 5th term of an arithmetic progression is 11 and the 9th term is 23. Find the first term, the common difference and the sum of the first 20 terms.

  1. Write the two conditions: a + 4d = 11 and a + 8d = 23.
  2. Subtract the first from the second: 4d = 12, so d = 3.
  3. Substitute into a + 4(3) = 11, giving a = 11 - 12 = -1.
  4. Apply S_n = n/2 [2a + (n - 1)d] with n = 20: S_20 = 10 [2(-1) + 19(3)] = 10 [-2 + 57] = 10 x 55.
  5. Check the terms: T_5 = -1 + 4(3) = 11 and T_9 = -1 + 8(3) = 23, both correct.

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

  1. Write the general term: T_(r+1) = 6Cr (2x)^(6-r) (-1/x^2)^r.
  2. Separate the powers of x: (2x)^(6-r) gives x^(6-r) and (-1/x^2)^r gives x^(-2r), so the total power of x is 6 - r - 2r = 6 - 3r.
  3. For the term independent of x, set 6 - 3r = 0, which gives r = 2.
  4. Substitute r = 2: T_3 = 6C2 x 2^(6-2) x (-1)^2 = 15 x 16 x 1.
  5. 6C2 = (6 x 5)/(2 x 1) = 15 and 2^4 = 16, so the term is 15 x 16 = 240.

The mistake to avoid

Using n instead of n - 1 in the AP and GP general terms, so that the 5th term comes out as a + 5d instead of a + 4d. The other standing error is quoting the sum to infinity for a GP with r greater than 1, which does not exist; check the value of r before you use that formula.

In the exam

Always state the general term formula in full before substituting, because the examiner awards a mark for the correct formula even when your arithmetic afterwards fails. In binomial questions, find r first from the power of x and only then work out the numerical coefficient, and do not forget to carry the sign of the negative term through the power of r.