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Binomial expansion

Expanding (1+x)^n and (a+f(x))^n for positive integer n; n! and nCr notation.

Spec MM2.4Section 1 Part 1: AS-level pure mathematics16 questions12 real past questions
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1Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

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

  1. A\(270\)
  2. B\(1080\)
  3. C\(2160\)
  4. D\(720\)
  5. E\(540\)
  6. F\(360\)
Worked solution

\(\binom{5}{3} 2^{2} (3x)^3 = 10 \times 4 \times 27 x^3 = 1080x^3\).

Full method — how and why
A single coefficient from \((a + bx)^n\). Use the general term rather than expanding the whole thing.
  1. Write the general term: \(\binom{5}{r} 2^{5-r}(3x)^r\).
    Every term of \((a+b)^n\) has the form \(\binom nr a^{n-r}b^r\); the powers of the two parts always add to \(n\).
  2. For \(x^3\) take \(r = 3\): \(\binom53 2^{2}(3x)^3\).
    The power of \(x\) comes only from the second bracket, so \(r\) is the power you want.
  3. Evaluate each factor: \(\binom53 = 10\), \(2^2 = 4\), \(3^3 = 27\).
    Keep them separate - combining too early is where slips happen.
  4. Multiply: \(10 \times 4 \times 27 = 1080\).
    \(10 \times 108 = 1080\).
Where it goes wrong
  • Forgetting the \(2^{2}\) and answering \(270\).
  • Using \(3^3 = 9\) or cubing only the \(x\), giving \(360\) - both are offered.
Takeaway The general term \(\binom nr a^{n-r}b^r\) carries three factors: the binomial coefficient and a power of each part.
Answer B · Binomial expansion (MM2.4) · question BIN-01
2Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

The coefficient of \(x^2\) in the expansion of \((1 + kx)^6\) is \(60\). Given that \(k > 0\), find \(k\).

  1. A\(2\)
  2. B\(4\)
  3. C\(3\)
  4. D\(\sqrt{10}\)
  5. E\(\dfrac{1}{2}\)
  6. F\(10\)
Worked solution

\(\binom62 k^2 = 15k^2 = 60\), so \(k^2 = 4\) and \(k = 2\).

Full method — how and why
A coefficient is given and a constant inside the bracket is unknown. Set up the coefficient equation and solve.
  1. General term of \((1 + kx)^6\): \(\binom6r (kx)^r\), so the \(x^2\) coefficient is \(\binom62 k^2\).
    The \(1^{6-r}\) factor is invisible, which is why this form is the easiest to work with.
  2. Compute \(\binom62 = 15\), so the equation is \(15k^2 = 60\).
    \(\binom62 = \frac{6 \times 5}{2} = 15\).
  3. Solve: \(k^2 = 4\), so \(k = \pm 2\); the condition \(k > 0\) gives \(k = 2\).
    Squaring means two roots; the stated condition selects one.
Where it goes wrong
  • Forgetting to square \(k\) and solving \(15k = 60\) to get \(k = 4\) - which is offered.
  • Using \(\binom62 = 30\) (that is \(6 \times 5\) without dividing by \(2\)).
Takeaway The \(x^r\) coefficient of \((1 + kx)^n\) is \(\binom nr k^r\) - the constant is raised to the same power as \(x\).
Answer A · Binomial expansion (MM2.4) · question BIN-02
3Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

What is the sum of the coefficients in the expansion of \((3x - 2)^{7}\)?

  1. A\(2187\)
  2. B\(-128\)
  3. C\(0\)
  4. D\(1\)
  5. E\(-1\)
  6. F\(128\)
Worked solution

Put \(x = 1\): \((3 - 2)^7 = 1\).

Full method — how and why
Sum of all coefficients: substitute \(x = 1\). No expansion needed at all.
  1. Note that setting \(x = 1\) makes every power of \(x\) equal to \(1\), so the expansion collapses to the sum of its coefficients.
    If \((3x-2)^7 = \sum c_k x^k\) then putting \(x = 1\) gives \(\sum c_k\).
  2. Evaluate the original expression at \(x = 1\): \((3 - 2)^7 = 1^7 = 1\).
    You never need a single coefficient.
Where it goes wrong
  • Expanding and trying to add eight terms under time pressure.
  • Substituting \(x = 0\), which gives only the constant term \((-2)^7 = -128\) - an offered option.
Takeaway \(f(1)\) is the sum of the coefficients; \(f(0)\) is the constant term; \(f(-1)\) is the alternating sum.
Answer D · Binomial expansion (MM2.4) · question BIN-04
4Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Evaluate \(\binom{10}{3} - \binom{10}{7} + \binom{9}{2}\).

  1. A\(36\)
  2. B\(240\)
  3. C\(0\)
  4. D\(84\)
  5. E\(120\)
  6. F\(156\)
Worked solution

\(\binom{10}{3} = \binom{10}{7}\), so the first two cancel; \(\binom92 = 36\).

Full method — how and why
A short calculation made trivial by the symmetry of binomial coefficients.
  1. Use the symmetry \(\binom nr = \binom n{n-r}\): since \(3 + 7 = 10\), \(\binom{10}{3} = \binom{10}{7}\).
    Choosing \(3\) objects from \(10\) is the same as choosing which \(7\) to leave behind.
  2. So the first two terms cancel, leaving \(\binom92\).
    Spotting the cancellation avoids computing \(120\) twice.
  3. Evaluate \(\binom92 = \frac{9 \times 8}{2} = 36\).
    \(\binom n2 = \frac{n(n-1)}{2}\).
Where it goes wrong
  • Computing all three coefficients and mis-adding, e.g. \(120 - 120 + 36\) done as \(156\).
  • Thinking \(\binom{10}{3} - \binom{10}{7}\) is \(\binom{10}{-4}\) or similar.
Takeaway \(\binom nr = \binom n{n-r}\): look for pairs whose lower entries add to \(n\).
Answer A · Binomial expansion (MM2.4) · question BIN-05
5Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

In the expansion of \((1 + x)^n\), the coefficients of \(x^4\) and \(x^5\) are equal. Find \(n\).

  1. A\(20\)
  2. B\(5\)
  3. C\(11\)
  4. D\(9\)
  5. E\(10\)
  6. F\(8\)
Worked solution

\(\binom n4 = \binom n5\) requires \(4 + 5 = n\) (by symmetry \(\binom nr = \binom n{n-r}\)), so \(n = 9\).

Full method — how and why
Two equal binomial coefficients from the same row. Symmetry gives the answer without computing factorials.
  1. The condition is \(\binom n4 = \binom n5\).
    In \((1+x)^n\) the coefficient of \(x^r\) is \(\binom nr\).
  2. Within one row of Pascal's triangle, two different entries are equal only when they are symmetric partners: \(\binom nr = \binom ns\) with \(r \neq s\) forces \(r + s = n\).
    The row rises to a single peak and falls symmetrically, so equal values are mirror images.
  3. Hence \(n = 4 + 5 = 9\).
    Check: \(\binom94 = 126 = \binom95\) ✓.
Where it goes wrong
  • Setting up the factorial equation and mis-cancelling to get \(n = 10\) or \(n = 8\).
  • Assuming \(n = 5\) because the coefficients 'stop' there.
Takeaway Equal entries in the same row of Pascal's triangle are symmetric partners: \(r + s = n\).
Answer D · Binomial expansion (MM2.4) · question BIN-06
6Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Expand \((1 - 2x)^4\) and hence find the coefficient of \(x^2\) in \((1 + 3x)(1 - 2x)^4\).

  1. A\(24\)
  2. B\(-24\)
  3. C\(48\)
  4. D\(0\)
  5. E\(12\)
  6. F\(-48\)
Worked solution

\((1-2x)^4 = 1 - 8x + 24x^2 - 32x^3 + 16x^4\). Coefficient of \(x^2\) in the product: \(1\cdot 24 + 3\cdot(-8) = 0\).

Full method — how and why
A product of a binomial expansion with a linear factor. Only two products can give \(x^2\), so collect just those.
  1. Expand \((1-2x)^4\): coefficients \(\binom4r(-2)^r\) give \(1 - 8x + 24x^2 - 32x^3 + 16x^4\).
    \(\binom41(-2) = -8\) and \(\binom42(-2)^2 = 6 \times 4 = 24\); remember to raise the \(-2\) to the power as well.
  2. To make \(x^2\) in the product, pair \(1 \times 24x^2\) and \(3x \times (-8x)\).
    Degrees must add to \(2\): \((0, 2)\) and \((1, 1)\) are the only options.
  3. Add: \(24 + (-24) = 0\).
    The two contributions cancel exactly - the answer is \(0\), not 'no term'.
Where it goes wrong
  • Taking the \(x^2\) coefficient of \((1-2x)^4\) as \(+24\) but forgetting the \(3x \times (-8x)\) contribution, giving \(24\).
  • Sign error making \((-2)^2 = -4\).
Takeaway For a coefficient in a product, list the degree pairs that add to the target and sum their contributions.
Answer D · Binomial expansion (MM2.4) · question BIN-07
7Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Find the coefficient of \(x^5\) in the expansion of \((1 + x)^{3}(1 + x)^{4}\).

  1. A\(35\)
  2. B\(21\)
  3. C\(12\)
  4. D\(1\)
  5. E\(28\)
  6. F\(7\)
Worked solution

\((1+x)^3(1+x)^4 = (1+x)^7\), and \(\binom75 = 21\).

Full method — how and why
Two binomials with the same bracket: combine the powers first, then use a single coefficient.
  1. Combine: \((1+x)^3(1+x)^4 = (1+x)^7\).
    Same base, so add the indices - this converts a product of expansions into one expansion.
  2. The coefficient of \(x^5\) is \(\binom75\).
    Standard general term with \(r = 5\).
  3. Evaluate using symmetry: \(\binom75 = \binom72 = \frac{7 \times 6}{2} = 21\).
    Computing \(\binom72\) is quicker than \(\binom75\) and gives the same value.
Where it goes wrong
  • Multiplying the separate expansions term by term and mis-collecting.
  • Multiplying the coefficients \(\binom35\) and \(\binom45\) - both of which are zero, and neither is relevant.
Takeaway Combine equal brackets before expanding; then one binomial coefficient does the whole job.
Answer B · Binomial expansion (MM2.4) · question BIN-10
8Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

Use the binomial expansion to find the value of \(1.01^{5}\) correct to \(4\) decimal places.

  1. A\(1.0515\)
  2. B\(1.0510\)
  3. C\(1.0505\)
  4. D\(1.0500\)
  5. E\(1.0410\)
  6. F\(1.0501\)
Worked solution

\((1 + 0.01)^5 = 1 + 5(0.01) + 10(0.0001) + 10(0.000001) + \ldots = 1 + 0.05 + 0.001 + 0.00001 + \ldots \approx 1.0510\).

Full method — how and why
An approximation with a small quantity. The binomial expansion converges fast because \(0.01\) is tiny.
  1. Write \(1.01^5 = (1 + 0.01)^5\) and expand: \(1 + 5(0.01) + 10(0.01)^2 + 10(0.01)^3 + \ldots\)
    Coefficients from row 5 of Pascal's triangle: \(1, 5, 10, 10, 5, 1\).
  2. Evaluate the terms: \(1 + 0.05 + 0.001 + 0.00001 + \ldots\)
    \((0.01)^2 = 0.0001\), so the third term is \(10 \times 0.0001 = 0.001\).
  3. Add as far as the 4-decimal-place accuracy requires: \(1.05101\ldots \approx 1.0510\).
    The fourth term only affects the fifth decimal place, so three terms suffice.
Where it goes wrong
  • Stopping after two terms and answering \(1.0500\).
  • Mis-scaling \((0.01)^2\) as \(0.001\), which gives \(1.0600\)-style errors.
Takeaway With a small \(x\), each binomial term is much smaller than the last - take just enough terms for the required accuracy.
Answer B · Binomial expansion (MM2.4) · question BIN-13
9Binomial expansion · spec MM2.4 · Paper 1 style · ★★☆

Find the term independent of \(x\) in the expansion of \(\left(x^2 + \dfrac{2}{x}\right)^6\).

  1. A\(60\)
  2. B\(15\)
  3. C\(160\)
  4. D\(480\)
  5. E\(240\)
  6. F\(120\)
Worked solution

General term \(\binom6r (x^2)^{6-r} \left(\frac2x\right)^r = \binom6r 2^r x^{12 - 3r}\). Need \(12 - 3r = 0\), \(r = 4\): \(\binom64 \times 16 = 15 \times 16 = 240\).

Full method — how and why
'Term independent of \(x\)' means the power of \(x\) is zero. Track the power of \(x\) in the general term and solve for \(r\).
  1. General term: \(\binom6r (x^2)^{6-r}\left(\frac2x\right)^r = \binom6r 2^r x^{12 - 2r - r}\).
    \((x^2)^{6-r}\) gives \(x^{12-2r}\) and \(\left(\frac2x\right)^r\) gives \(2^r x^{-r}\); combine the powers.
  2. So the power of \(x\) is \(12 - 3r\).
    Getting this expression right is the whole question.
  3. Set it to zero: \(12 - 3r = 0\), so \(r = 4\).
    'Independent of \(x\)' means \(x^0\).
  4. Evaluate the term: \(\binom64 \times 2^4 = 15 \times 16 = 240\).
    \(\binom64 = \binom62 = 15\) by symmetry.
Where it goes wrong
  • Using the power \(12 - 2r\) and getting \(r = 6\), giving \(2^6 = 64\).
  • Forgetting the \(2^r\) factor and answering \(15\).
Takeaway Write the power of \(x\) in the general term as a linear expression in \(r\), then set it to the value you need.
Answer E · Binomial expansion (MM2.4) · question BIN-03
10Binomial expansion · spec MM2.4 · Paper 1 style · ★★☆

The first three terms in the expansion of \((1 + ax)^n\) in ascending powers of \(x\) are \(1 + 12x + 54x^2\). Find \(n\).

  1. A\(9\)
  2. B\(6\)
  3. C\(10\)
  4. D\(4\)
  5. E\(12\)
  6. F\(8\)
Worked solution

\(na = 12\) and \(\frac{n(n-1)}{2}a^2 = 54\). Dividing the second by the square of the first: \(\frac{n-1}{2n} = \frac{54}{144} = \frac38\), so \(8n - 8 = 6n\) and \(n = 4\) (then \(a = 3\); check \(\binom42 \cdot 9 = 54\)).

Full method — how and why
Two coefficients, two unknowns (\(a\) and \(n\)). Dividing the equations cleverly eliminates \(a\).
  1. Write the first two coefficients: \(\binom n1 a = na = 12\) and \(\binom n2 a^2 = \frac{n(n-1)}{2}a^2 = 54\).
    The expansion of \((1+ax)^n\) starts \(1 + nax + \frac{n(n-1)}{2}a^2x^2\).
  2. Square the first equation: \(n^2a^2 = 144\).
    Squaring makes the power of \(a\) match the second equation, so dividing will cancel it.
  3. Divide the second by this: \(\dfrac{n(n-1)a^2/2}{n^2a^2} = \dfrac{n-1}{2n} = \dfrac{54}{144} = \dfrac38\).
    Both \(a^2\) and one factor of \(n\) cancel.
  4. Cross-multiply: \(8(n-1) = 6n \Rightarrow 8n - 8 = 6n \Rightarrow n = 4\).
    Then \(a = 3\); check \(\binom42 \times 9 = 6 \times 9 = 54\) ✓.
Where it goes wrong
  • Dividing the second equation by the first (not its square), which leaves an \(a\) behind and stalls.
  • Using \(\binom n2 = \frac{n^2}{2}\).
Takeaway To eliminate a constant that appears to different powers, raise one equation to the matching power first, then divide.
Answer D · Binomial expansion (MM2.4) · question BIN-08
11Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

Which of the following statements is/are true?

  1. I\(\binom{n}{r} = \binom{n}{n-r}\) for all integers \(0 \leq r \leq n\)
  2. II\(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\) for all integers \(0 \leq r < n\)
  3. III\(\displaystyle\sum_{r=0}^{n} \binom{n}{r} = 2^{n}\) for all positive integers \(n\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is the symmetry of Pascal's triangle; II is Pascal's rule (each entry is the sum of the two above it); III is \((1+1)^n\) expanded. All true.

Full method — how and why
Three standard properties of binomial coefficients. Each has a one-line reason and a one-line check.
  1. I: \(\binom nr = \binom n{n-r}\) - choosing \(r\) objects is the same as choosing the \(n-r\) to leave. True.
    Visible as the left-right symmetry of each row of Pascal's triangle.
  2. II: \(\binom nr + \binom n{r+1} = \binom{n+1}{r+1}\) - Pascal's rule, i.e. each entry is the sum of the two above it. True.
    Check with numbers: \(\binom32 + \binom33 = 3 + 1 = 4 = \binom43\) ✓.
  3. III: \(\sum_{r=0}^n \binom nr = 2^n\) - put \(x = 1\) in \((1+x)^n\). True.
    Equivalently, a set with \(n\) elements has \(2^n\) subsets, counted by size.
  4. All three: 'I, II and III'.
    Match to the options.
Where it goes wrong
  • Doubting Pascal's rule because of the index shift on the right - test it with small numbers.
  • Thinking the sum in III is \(2^{n+1}\) by miscounting the \(n+1\) terms.
Takeaway Symmetry, Pascal's rule and row sum \(2^n\) are the three facts about \(\binom nr\) worth knowing cold.
Answer H · Binomial expansion (MM2.4) · question BIN-09
12Binomial expansion · spec MM2.4 · Paper 1 style · ★★☆

Find the value of \((\sqrt{2} + 1)^4 + (\sqrt{2} - 1)^4\).

  1. A\(34\)
  2. B\(36\)
  3. C\(17\)
  4. D\(34 + 24\sqrt2\)
  5. E\(24\sqrt{2}\)
  6. F\(16\)
Worked solution

Odd powers of \(\sqrt2\) cancel: \(2\left[(\sqrt2)^4 + \binom42(\sqrt2)^2 + 1\right] = 2[4 + 12 + 1] = 34\).

Full method — how and why
Adding conjugate expansions. The odd-power terms cancel, so you only need the even ones - and the surds disappear.
  1. Expand both with the binomial theorem. Their terms are identical except that odd powers of \(1\) and \(-1\) alternate in sign.
    \((\sqrt2 + 1)^4\) and \((\sqrt2 - 1)^4\) differ only in the sign of the odd-numbered terms.
  2. Adding doubles the even-power terms and cancels the odd ones: \(2\left[(\sqrt2)^4 + \binom42(\sqrt2)^2 + 1\right]\).
    The terms kept are \(r = 0, 2, 4\) in \(\binom4r(\sqrt2)^{4-r}\) - exactly those with an even power of \(\sqrt2\), which are rational.
  3. Evaluate: \((\sqrt2)^4 = 4\), \(\binom42(\sqrt2)^2 = 6 \times 2 = 12\), and \(1\).
    \((\sqrt2)^2 = 2\) and \((\sqrt2)^4 = 4\).
  4. Total: \(2(4 + 12 + 1) = 2 \times 17 = 34\).
    Rational, as expected - a sure sign the cancellation was done correctly.
Where it goes wrong
  • Forgetting the factor \(2\) and answering \(17\).
  • Keeping a surd term and choosing \(34 + 24\sqrt2\), which is \((\sqrt2+1)^4\) alone.
Takeaway Adding conjugate binomials kills the odd terms, leaving a rational answer - expand only the even ones.
Answer A · Binomial expansion (MM2.4) · question BIN-11
13Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

The expansion of \(\left(2x - \dfrac{1}{x^2}\right)^{9}\) contains a term in \(x^{k}\) for each of which of the following values of \(k\)?

  1. I\(k = 0\)
  2. II\(k = 3\)
  3. III\(k = -6\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

General term: \(\binom9r (2x)^{9-r}(-x^{-2})^r\) has power \(9 - r - 2r = 9 - 3r\), which takes values \(9, 6, 3, 0, -3, \ldots, -18\). \(k = 0\) needs \(r = 3\): yes. \(k = 3\) needs \(r = 2\): yes. \(k = -6\) needs \(r = 5\): yes. All three.

Full method — how and why
Which powers of \(x\) actually occur. Find the general power as an expression in \(r\) and see which target values it can hit.
  1. General term: \(\binom9r (2x)^{9-r}\left(-x^{-2}\right)^r\), whose power of \(x\) is \((9 - r) + (-2r) = 9 - 3r\).
    Add the powers contributed by each bracket; the \(-\frac{1}{x^2}\) contributes \(-2\) per copy.
  2. As \(r\) runs from \(0\) to \(9\), the power takes the values \(9, 6, 3, 0, -3, -6, -9, -12, -15, -18\).
    They step down by \(3\) - so only multiples of \(3\) appear, all the way down to \(-18\).
  3. \(k = 0\) needs \(r = 3\) ✓; \(k = 3\) needs \(r = 2\) ✓; \(k = -6\) needs \(r = 5\) ✓.
    Each value of \(r\) must be a whole number between \(0\) and \(9\) - that is the only restriction.
  4. All three occur: 'I, II and III'.
    Solve \(9 - 3r = k\) for each target and check that \(r\) is a valid index.
Where it goes wrong
  • Forgetting that the second bracket contributes \(x^{-2r}\), not \(x^{-r}\).
  • Assuming negative powers cannot appear because it is a 'binomial expansion'.
Takeaway Express the power of \(x\) as a linear function of \(r\); a term exists exactly when the required \(r\) is an integer in range.
Answer H · Binomial expansion (MM2.4) · question BIN-12
14Binomial expansion · spec MM2.4 · Paper 1 style · ★★☆

The coefficient of \(x^2\) in \((1 + x)^n + (1 + x)^{n+1}\) is \(36\). Find \(n\).

  1. A\(4\)
  2. B\(8\)
  3. C\(6\)
  4. D\(5\)
  5. E\(9\)
  6. F\(7\)
Worked solution

\(\binom n2 + \binom{n+1}2 = \frac{n(n-1) + (n+1)n}{2} = n^2 = 36\), so \(n = 6\).

Full method — how and why
Two expansions added. Write both coefficients in factorial form and the sum simplifies remarkably.
  1. The \(x^2\) coefficients are \(\binom n2\) and \(\binom{n+1}{2}\).
    One from each expansion; the total coefficient of \(x^2\) is their sum.
  2. Write them out: \(\dfrac{n(n-1)}{2} + \dfrac{(n+1)n}{2} = \dfrac{n\left[(n-1) + (n+1)\right]}{2}\).
    Factor out \(\frac n2\) - the brackets then collapse.
  3. Simplify: \(\dfrac{n \times 2n}{2} = n^2\).
    A surprisingly clean result; recognising it makes the final step trivial.
  4. Solve \(n^2 = 36\), so \(n = 6\) (positive integer).
    \(n\) counts an index, so the negative root is rejected.
Where it goes wrong
  • Using Pascal's rule to write the sum as \(\binom{n+1}{3}\) - that rule adds adjacent entries in the same row, not the same entry in different rows.
  • Taking \(n = \pm 6\) and offering \(-6\).
Takeaway \(\binom n2 + \binom{n+1}{2} = n^2\): write coefficients in factorial form and look for common factors before solving.
Answer C · Binomial expansion (MM2.4) · question BIN-15
15Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

\(n\) is a positive integer. Which of the following statements is/are true?

  1. IThe coefficient of \(x^n\) in \((1 + x)^{2n}\) is \(\binom{2n}{n}\).
  2. IIThe sum of the coefficients in the expansion of \((1 - x)^n\) is \(0\).
  3. III\((1 + x)^n \geq 1 + nx\) for all \(x \geq 0\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is the general term with \(r = n\). II: put \(x = 1\). III: expanding \((1 + x)^n\) gives \(1 + nx\) plus non-negative terms when \(x \geq 0\).

Full method — how and why
Three binomial facts: the general term, substituting a value, and a term-by-term inequality.
  1. I: the coefficient of \(x^r\) in \((1+x)^m\) is \(\binom mr\). With \(m = 2n\) and \(r = n\) that is \(\binom{2n}{n}\). True.
    Straight from the binomial theorem.
  2. II: the sum of the coefficients of any polynomial is its value at \(x=1\). Here \((1-1)^n = 0^n = 0\) for \(n \geq 1\). True.
    Substituting \(x=1\) makes every power equal \(1\), so the value is the coefficient sum.
  3. III: for \(x \geq 0\), \((1+x)^n = 1 + nx + \binom n2 x^2 + \cdots\), and every omitted term is non-negative. So \((1+x)^n \geq 1+nx\). True.
    All binomial coefficients and all powers of a non-negative \(x\) are non-negative.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Thinking II fails for \(n=0\) - the question says \(n\) is a positive integer.
  • Doubting III for large \(n\); the extra terms only help, since \(x \geq 0\).
  • Confusing \(\binom{2n}{n}\) with \(\binom{2n}{2}\) or \(2\binom nn\).
Takeaway Sum of coefficients \(=\) value at \(x=1\); and \((1+x)^n \geq 1+nx\) for \(x \geq 0\) (Bernoulli) follows from discarding non-negative terms.
Answer H · Binomial expansion (MM2.4) · question X2-13
16Binomial expansion · spec MM2.4 · Paper 1 style · ★★★

How many terms in the expansion of \((x + y + z)^{4}\) (after collecting like terms) are there?

  1. A\(16\)
  2. B\(15\)
  3. C\(10\)
  4. D\(5\)
  5. E\(12\)
  6. F\(81\)
Worked solution

Terms are \(x^ay^bz^c\) with \(a + b + c = 4\), \(a, b, c \geq 0\). Counting: for \(a = 0\) there are \(5\) choices of \((b, c)\), \(a = 1\): \(4\), ..., \(a = 4\): \(1\). Total \(5 + 4 + 3 + 2 + 1 = 15\).

Full method — how and why
A trinomial expansion. Counting the distinct terms is a counting problem: how many ways can the powers add to \(4\)?
  1. Every term has the form \(x^ay^bz^c\) with \(a + b + c = 4\) and \(a, b, c \geq 0\).
    Each of the four brackets contributes exactly one of \(x\), \(y\) or \(z\), so the powers add to \(4\).
  2. Count systematically by the value of \(a\): if \(a = 0\) then \(b + c = 4\), giving \(5\) pairs; \(a = 1\) gives \(4\); \(a=2\) gives \(3\); \(a=3\) gives \(2\); \(a=4\) gives \(1\).
    For a fixed \(a\), the number of \((b,c)\) pairs is \((4 - a) + 1\).
  3. Total: \(5 + 4 + 3 + 2 + 1 = 15\).
    A triangular number - the standard result \(\binom{n+2}{2}\) for three variables, here \(\binom62 = 15\).
Where it goes wrong
  • Answering \(3^4 = 81\) - that counts the terms before collecting like terms.
  • Answering \(5\) by counting only the powers of \(x\).
Takeaway Distinct terms of \((x+y+z)^n\) correspond to solutions of \(a+b+c=n\) in non-negative integers: \(\binom{n+2}{2}\) of them.
Answer B · Binomial expansion (MM2.4) · question BIN-14

How this topic has been examined

Real TMUA questions on Binomial expansion, newest first. Links open the official paper at the question and the online marking page.

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