IGCSE Additional Mathematics: Series and the Binomial Theorem Practice Questions
An arithmetic progression adds a common difference each term, a geometric progression multiplies by a common ratio, and the binomial theorem expands a bracket raised to a power using combinations as coefficients.
Topic 13 of Cambridge IGCSE Additional Mathematics 0606 covers three related tools. Finding a single term of a binomial expansion, without expanding the whole thing, is the technique worth mastering. The questions below cover it alongside both progressions.
What you need to know for Series and the Binomial Theorem
- Binomial theoremThe general term of (a + b)n is n choose r, multiplied by a to the power (n minus r), multiplied by b to the power r.
- Finding one termSet the index of the required variable equal to r, then evaluate the general term. Expanding everything wastes time.
- Arithmetic progressionThe nth term is a + (n minus 1)d. The sum of n terms is n over 2, multiplied by [2a + (n minus 1)d].
- Geometric progressionThe nth term is arn-1. The sum of n terms is a(rn minus 1) divided by (r minus 1).
- Sum to infinityExists only when the size of r is less than 1, and equals a divided by (1 minus r).
IGCSE Additional Mathematics Series and the Binomial Theorem questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Find the coefficient of x3 in the expansion of (2 + x)5.
Show the worked answer
- The general term is 5 choose r, multiplied by 2 to the power (5 minus r), multiplied by x to the power r.
- For the x3 term, set r = 3.
- 5 choose 3 = 10, and 2 to the power (5 minus 3) = 22 = 4.
- The coefficient is 10 x 4 = 40, so the term is 40x3.
An arithmetic progression has first term 5 and common difference 3. Calculate the sum of the first 20 terms.
Show the worked answer
- Use the sum formula: Sn = n over 2, multiplied by [2a + (n minus 1)d].
- Substitute a = 5, d = 3 and n = 20: S = 10 x [10 + 19 x 3].
- 19 x 3 = 57, so the bracket is 10 + 57 = 67.
- S = 10 x 67 = 670.
A geometric progression has first term 3 and common ratio 2. Find the 8th term and the sum of the first 8 terms.
Show the worked answer
- The nth term is arn-1, so the 8th term is 3 x 27.
- 27 = 128, so the 8th term is 3 x 128 = 384.
- The sum is a(rn minus 1) divided by (r minus 1) = 3(28 minus 1) divided by 1.
- 28 = 256, so the sum is 3 x 255 = 765.
A geometric progression has first term 16 and common ratio 0.5. Explain why a sum to infinity exists and calculate it.
Show the worked answer
- A sum to infinity exists only when the size of the common ratio is less than 1.
- Here r = 0.5, whose size is less than 1, so the terms decrease towards zero and the sum converges.
- Sum to infinity = a divided by (1 minus r) = 16 divided by (1 minus 0.5).
- = 16 divided by 0.5 = 32.
Common mistakes in this topic
- Omitting the power of the constant term in a binomial expansion.
- Using n instead of n minus 1 in progression formulas.
- Applying the sum to infinity formula without checking the convergence condition.
- Confusing the arithmetic and geometric sum formulas.
- Expanding an entire binomial when only one term is required.
Exam tips
- Write the general term before substituting. It is a method mark and prevents index errors.
- For a specific term, set r equal to the required index and stop there.
- Both progression nth term formulas use n minus 1. Remember it once and it covers both.
- State the convergence condition explicitly in any sum to infinity question.
- Sense check a geometric sum: with r greater than 1 the sum grows fast, with r less than 1 it approaches a limit.
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Series and the Binomial Theorem FAQs
How do I find one term of a binomial expansion?
Use the general term: n choose r, multiplied by a to the power n minus r, multiplied by b to the power r. Set r equal to the index of the variable you want, then evaluate. There is no need to expand the whole bracket.
What is the sum formula for an arithmetic progression?
The sum of n terms equals n over 2, multiplied by the bracket 2a plus n minus 1 times d, where a is the first term and d the common difference. The n minus 1 reflects that reaching the nth term takes one fewer step than the term number.
When does a geometric series have a sum to infinity?
Only when the size of the common ratio is less than 1, so the terms shrink towards zero and the total converges. The sum to infinity is then the first term divided by one minus the common ratio. Always state the condition before applying the formula.
Why is the index n minus 1 in the nth term formulas?
The first term requires no steps, the second requires one, and so on, so reaching the nth term takes n minus 1 steps. That is why the arithmetic formula multiplies d by n minus 1 and the geometric formula raises r to the power n minus 1.
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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.