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DSE Maths: Mathematical Induction and the Binomial Theorem Prep Guide (for tutors) | Three steps, general term, coefficients
Faced with prove for all positive integers, students go blank; faced with a binomial, they multiply everything out. This lesson turns both into a grid: three boxes for induction, one general term for the binomial.
Objectives — Students can: (1) write all three induction steps and say why a missing step is not a proof; (2) substitute the induction hypothesis to handle divisibility claims; (3) use the general term C(n,r) a to the n minus r times b to the r to find a required coefficient, the constant term and the sum of coefficients.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Index laws | How to split 9 to the k plus one | 9 times 9 to the k |
| Combinations | How C(n,r) is computed | C(7,3) is 35 |
| Polynomials | Collecting like terms | Add the coefficients |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-12 min | Three steps | P(1), assume P(k), prove P(k+1) | Write all three boxes |
| 12-24 min | Sum example | Add the last term to the hypothesis | Get 5050 |
| 24-36 min | Divisibility example | Factor out 9 to the k and substitute | Write it as 8 times an integer |
| 36-48 min | General term | C(n,r) a to the n minus r b to the r | Coefficient of x cubed is 280 |
| 48-56 min | Constant term and coefficient sum | Set the index to zero; put x equal to 1 | Get 20 and 2,187 |
| 56-60 min | Exit check | Collect three questions and mark live | All three right |
03 Teaching script (read it out)
Scene 1: Induction is a three-box grid, not an essay
Step one: verify P(1). Step two: assume P(k). Step three: use that assumption to prove P(k+1).
Example: prove that 1 up to n equals n(n+1)/2. At n equal to 1 the left side is 1 and the right side is 1 times 2 over 2, which is 1.
Assume 1 up to k equals k(k+1)/2; then 1 up to k+1 equals k(k+1)/2 plus (k+1), which is (k+1)(k+2)/2, exactly the formula with n replaced by k+1.
Hence 1 up to 100 equals 5050.
The usual lost mark: step two says assume P(n) and step three proves P(n) again, which is circular.
Example: prove that 1 up to n equals n(n+1)/2. At n equal to 1 the left side is 1 and the right side is 1 times 2 over 2, which is 1.
Assume 1 up to k equals k(k+1)/2; then 1 up to k+1 equals k(k+1)/2 plus (k+1), which is (k+1)(k+2)/2, exactly the formula with n replaced by k+1.
Hence 1 up to 100 equals 5050.
The usual lost mark: step two says assume P(n) and step three proves P(n) again, which is circular.
Ask the student: which letter goes in the assumption and which in the conclusion (k, then k+1).
Scene 2: Divisibility: swap the hypothesis in as one block
Prove that 9 to the n minus 1 is divisible by 8. At n equal to 1, 9 minus 1 is 8, which is 8 times 1.
Assume 9 to the k minus 1 equals 8m. Then 9 to the k plus one minus 1 equals 9 times 9 to the k minus 1; replacing 9 to the k by 8m plus 1 gives 9(8m plus 1) minus 1, which is 72m plus 8, that is 8(9m plus 1).
Since 9m plus 1 is an integer, the claim holds at k+1.
Check n equal to 3: 729 minus 1 is 728, which is 8 times 91.
One technique only: replace the whole 9 to the k block and never expand it.
Assume 9 to the k minus 1 equals 8m. Then 9 to the k plus one minus 1 equals 9 times 9 to the k minus 1; replacing 9 to the k by 8m plus 1 gives 9(8m plus 1) minus 1, which is 72m plus 8, that is 8(9m plus 1).
Since 9m plus 1 is an integer, the claim holds at k+1.
Check n equal to 3: 729 minus 1 is 728, which is 8 times 91.
One technique only: replace the whole 9 to the k block and never expand it.
Ask the student: how do you split 9 to the k plus one (9 times 9 to the k).
Scene 3: Binomial: write the general term, never multiply out
The general term is C(n,r) times a to the n minus r times b to the r.
For (1 plus 2x) to the 7, the coefficient of x cubed is C(7,3) times 2 cubed, that is 35 times 8, equal to 280; the coefficient of x squared is C(7,2) times 4, equal to 84.
For (x plus 1 over x) to the 6, the constant term needs the index of x to be zero, so r is 3 and the coefficient is C(6,3), equal to 20.
The sum of coefficients: put x equal to 1, giving 3 to the 7, equal to 2,187.
Coefficient and term are different answers: the coefficient of x cubed is 280, the term is 280x cubed.
For (1 plus 2x) to the 7, the coefficient of x cubed is C(7,3) times 2 cubed, that is 35 times 8, equal to 280; the coefficient of x squared is C(7,2) times 4, equal to 84.
For (x plus 1 over x) to the 6, the constant term needs the index of x to be zero, so r is 3 and the coefficient is C(6,3), equal to 20.
The sum of coefficients: put x equal to 1, giving 3 to the 7, equal to 2,187.
Coefficient and term are different answers: the coefficient of x cubed is 280, the term is 280x cubed.
Ask the student: which equation fixes r for a constant term (the index equals zero).
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Proof | Three-step induction | P(1), assume P(k), prove P(k+1) | All three steps carry marks |
| Proof | Divisibility | Write the hypothesis as 8m and substitute | End as 8 times an integer |
| Calculation | Coefficient of a given power | General term C(n,r) a to the n minus r b to the r | Coefficient of x cubed is 280 |
| Calculation | Constant term and coefficient sum | Set the index to zero; put x equal to 1 | 20 and 2,187 |
Sanity check (teach this): All three induction steps are needed, especially P(1): without a base case the rest is circular.
Keep the letters apart, k in the assumption and k+1 in the conclusion; the same letter on both sides means the answer was copied.
Binomial coefficients are symmetric: C(7,3) equals C(7,4) equals 35, so an asymmetric pair signals an error.
The constant term is not always in the middle: the r you solve must lie between 0 and n, and r above n means there is no constant term.
Keep the letters apart, k in the assumption and k+1 in the conclusion; the same letter on both sides means the answer was copied.
Binomial coefficients are symmetric: C(7,3) equals C(7,4) equals 35, so an asymmetric pair signals an error.
The constant term is not always in the middle: the r you solve must lie between 0 and n, and r above n means there is no constant term.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Sum from 1 to 100 | 5050 | n(n+1) over 2 |
| Is 9 cubed minus 1 divisible by 8 | 728, which is 8 times 91 | A base-case check |
| Coefficient of x cubed in (1 plus 2x) to the 7 | 280 | C(7,3) times 8 |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Same letter in assumption and conclusion | Assume P(n), prove P(n) | Assume for k and prove k+1; the letters must differ. |
| Skipping the base case | The assumption is enough | Without P(1) there is no start, so the chain breaks and scores zero. |
| Giving the term instead of the coefficient | The answer is 280x cubed when the coefficient was asked | A coefficient is 280; the term is 280x cubed. |
| Setting r equal to zero for a constant term | The constant term is always the first term | Solve the index equal to zero; for (x plus 1 over x) to the 6 that gives r equal to 3. |
07 Homework (paper, section and question numbers)
- Practice: writing the three induction steps — Fill the three boxes in order
- Practice: general term and coefficients — Write the general term before substituting
- All lesson-prep guides — Compare the ten-section structure
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Sum from 1 to 100 | 5050 | All three right: move to advanced counting |
| Coefficient of x cubed in (1 plus 2x) to the 7 | 280 | One wrong: rewrite the general term |
| Constant term of (x plus 1 over x) to the 6 | 20 | Two wrong: back to act three |
09 Tutor FAQ
Does induction need words, not just algebra?
Yes, especially the line that the assumption gives the k+1 case; a bare calculation without that conclusion loses marks.
Must students memorise binomial expansions?
No, the general term is enough; expansions are only a check for small n.
Why put x equal to 1 for the coefficient sum?
Every x becomes 1, so what remains is exactly the sum of all coefficients.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
