MathsISC Class 11 · Binomial Theorem

Middle Terms, Binomial Coefficients, Approximations and Divisibility

Updated 17 October 2026 · Fact-checked

In (a + b)ⁿ the middle term is T₍ₙ/₂₎₊₁ for even n and the two middle terms are T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂ for odd n. The sum of all binomial coefficients is 2ⁿ; the sums of the odd-placed and even-placed coefficients are each 2ⁿ⁻¹. Powers can be approximated and divisibility proved by splitting a number as (1 + small)ⁿ.

Understand Middle Terms, Binomial Coefficients and Applications

Middle terms. If n is even, there is one middle term, the $(\frac{n}{2} + 1)$ th. If n is odd, there are two middle terms, the $\frac{n + 1}{2}$ th and $\frac{n + 3}{2}$ th. The middle term has the greatest binomial coefficient: $^{n} C_{n /2}$ for even n and $^{n} C_{(n - 1) /2} =^{n} C_{(n + 1) /2}$ for odd n.

Sums of coefficients. Putting a = b = 1: $^{n} C_{0} +^{n} C_{1} + ... +^{n} C_{n} = 2^{n}$ . Putting a = 1, b = −1: $^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - ... = 0$ , so the sum of coefficients of the even-placed terms equals that of the odd-placed terms, each $2^{n - 1}$ . Also $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .

Approximation. For small x, $(1 + x)^{n} \approx 1 + n x + \frac{n ( n - 1 )}{2} x^{2}$ . Taking the first two or three terms gives a value to the required number of decimal places.

Divisibility. To prove that $9^{n + 1} - 8 n - 9$ is divisible by 64, write $9^{n + 1} = (1 + 8)^{n + 1} = 1 + 8 (n + 1) + 8^{2} \cdot (terms)$ . Then $9^{n + 1} - 8 n - 9 = 64 \times (an integer)$ . The same idea shows $7^{n} - 1$ is divisible by 6 and $1 1^{n} - 1$ by 10.

Strategy. Split the number as (a + b) where a or b is small or a multiple of the divisor, expand, and pick out the terms that are multiples of the divisor.

Middle Terms, Binomial Coefficients and Applications: formulas to remember

Middle terms: $n even: $T_{n/2 + 1}$ | n odd: $T_{(n+1)/2}$ and $T_{(n+3)/2}$$; The middle term has the greatest coefficient.
Sum of coefficients: $\sum ^nC_r = 2^n$ | odd-placed sum = even-placed sum = $2^{n-1}$; From $(1 + 1)^{n}$ and $(1 - 1)^{n}$ .
Approximation: $(1 + x)^n ≈ 1 + nx + \frac{n(n-1)}{2}x^2$; For small |x|.
Divisibility trick: $(1 + 8)^{n+1} = 1 + 8(n + 1) + 64 k$; Isolates the multiples of 64.

Middle Terms, Binomial Coefficients and Applications in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Greatest coefficient of a seating pattern

Suppose the number of ways of choosing r seats from 10 is given by the binomial coefficients $^{10} C_{r}$ .

Find the greatest value, the sum of all values and the sum for an odd number of seats chosen.

Try it first, then show the solutionSolution

The greatest coefficient is the middle one, $^{10} C_{5} = 252$ .
The sum of all coefficients is $2^{10} = 1024$ .
The sum of the coefficients with odd r is $2^{9} = 512$ .

Answer: 252; 1024; 512.

Memory hook: The middle coefficient is the largest.

Tricks to remember Middle Terms, Binomial Coefficients and Applications

Short memory aids. Say each one aloud once, then use it on a question the same day.

Even n: one; odd n: two
Number of middle terms depends on whether n is even or odd.
Use it when: Counting middle terms.
Write 99 as 100 − 1
Split the number so that powers of a small or divisible number appear.
Use it when: Approximations and divisibility.

Middle Terms, Binomial Coefficients and Applications: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Sum of binomial coefficients · 1 mark · MCQBoard level

The sum of all binomial coefficients in the expansion of $(1 + x)^{8}$ is:
(a) 64
(b) 128
(c) 256
(d) 512

Try it first, then show the solutionSolution

$2^{8} = 256$ .

Answer: (c) 256

Example 2: Middle term · 3 marks · Short answerBoard level

Find the middle term in the expansion of $(x + \frac{1}{x})^{10}$ .

Try it first, then show the solutionSolution

n = 10 is even, so the middle term is $T_{6}$ . (1 mark)
$T_{6} =^{10} C_{5} x^{5} (\frac{1}{x})^{5} =^{10} C_{5} \cdot x^{0}$ . (1 mark)
$= 252$ . (1 mark)

Answer: 252

Example 3: Two middle terms · 3 marks · Short answerHarder board level

Find the middle terms of $(x - \frac{1}{x})^{7}$ .

Try it first, then show the solutionSolution

n = 7 is odd, so the middle terms are $T_{4}$ and $T_{5}$ . (1 mark)
$T_{4} =^{7} C_{3} x^{4} (- \frac{1}{x})^{3} = - 35 x$ . (1 mark)
$T_{5} =^{7} C_{4} x^{3} (\frac{1}{x})^{4} = \frac{35}{x}$ . (1 mark)

Answer: −35x and 35/x.

Example 4: Approximation · 4 marks · Short answerHarder board level

Evaluate $(1.02)^{6}$ correct to four decimal places and find $(0.99)^{5}$ to three decimals.

Try it first, then show the solutionSolution

$(1.02)^{6} = 1 + 6 (0.02) + 15 (0.0004) + 20 (0.000008) + ...$ . (1 mark)
$= 1 + 0.12 + 0.006 + 0.00016 + 0.0000024 = 1.1261624 \approx 1.1262$ . (1.5 marks)
$(0.99)^{5} = (1 - 0.01)^{5} = 1 - 0.05 + 10 (0.0001) - 10 (0.000001) + ...$ . (1 mark)
$= 1 - 0.05 + 0.001 - 0.00001 = 0.95099 \approx 0.951$ . (0.5 mark)

Answer: 1.1262; 0.951.

Example 5: Divisibility · 5 marks · Long answerHOTS / competency

Prove that $3^{2 n + 2} - 8 n - 9$ is divisible by 64 for every positive integer n. Find also the sum of the coefficients of the even-placed terms in $(1 + x)^{10}$ .

Try it first, then show the solutionSolution

$3^{2 n + 2} = 9^{n + 1} = (1 + 8)^{n + 1}$ . (1 mark)
$= 1 + 8 (n + 1) +^{n + 1} C_{2} 8^{2} +^{n + 1} C_{3} 8^{3} + ...$ . (1.5 marks)
So $9^{n + 1} - 8 n - 9 = 64 [^{n + 1} C_{2} + 8 \cdot^{n + 1} C_{3} + ...]$ , a multiple of 64. (1.5 marks)
The sum of even-placed coefficients in $(1 + x)^{10}$ is $2^{9} = 512$ . (1 mark)

Answer: Divisible by 64; 512.

Where students lose marks in Middle Terms, Binomial Coefficients and Applications

The slips examiners see most often, and what to write instead.

Mistake: Taking T₅ as the middle term of n = 9
Ignoring that n is odd.
Do this: n = 9 has two middle terms T₅ and T₆.
Usually costs: 1 to 2 marks.
Mistake: Dropping the sign of b
Writing the middle term of $(x - 1/ x)^{7}$ with the wrong sign.
Do this: Keep (−1)^r in the term.
Usually costs: 1 mark.
Mistake: Not showing the multiple of the divisor
Concluding divisibility without expansion.
Do this: Show that all remaining terms contain the factor 64.
Usually costs: 1 to 2 marks.

Exam tips

Decide first how many middle terms there are.
Write the series terms to the required decimal accuracy.
In divisibility proofs, write the number as 1 + multiple of the divisor.
Use 2ⁿ and 2ⁿ⁻¹ for sums of coefficients.

Practice questions on Middle Terms, Binomial Coefficients and Applications

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of middle terms in the expansion of $(a + b)^{9}$ is:
(a) 1
(b) 2
(c) 3
(d) 0
Hint
n is odd.
Answer
(b) 2
Q2 · 2 marks · Short answerBoard level
Find the sum of the coefficients in the expansion of $(2 x + 3 y)^{5}$ .
Hint
Put x = y = 1.
Answer
$(2 + 3)^{5} = 3125$ .
Q3 · 3 marks · Short answerHarder board level
Find the middle term of $(2 x - \frac{y}{3})^{8}$ .
Hint
T5.
Answer
$T_{5} =^{8} C_{4} (2 x)^{4} (- \frac{y}{3})^{4} = 70 \times 16 x^{4} \times \frac{y ^{4}}{81} = \frac{1120}{81} x^{4} y^{4}$ .
Q4 · 3 marks · Short answerHarder board level
Using the binomial theorem, show that $1 1^{n} - 1$ is divisible by 10.
Hint
11 = 1 + 10.
Answer
$1 1^{n} = (1 + 10)^{n} = 1 + 10 n + 1 0^{2} (...)$ , so $1 1^{n} - 1 = 10 [n + 10 (...)]$ .
Q5 · 3 marks · Short answerHarder board level
Find the value of $^{10} C_{0} +^{10} C_{2} +^{10} C_{4} +^{10} C_{6} +^{10} C_{8} +^{10} C_{10}$ .
Hint
Sum of even-placed coefficients.
Answer
$2^{9} = 512$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a game designer needs the number of subsets of a 12-element set having an even number of elements, and the largest group of equal-sized subsets.
(i) State the number of subsets of size 6. (1 mark)
(ii) State the number of subsets of even size. (1 mark)
(iii) Find which size has the most subsets. (1 mark)
(iv) State the total number of subsets. (1 mark)
Hint
Binomial coefficients $^{12} C_{r}$ .
Answer
(i) $^{12} C_{6} = 924$ . (ii) $2^{11} = 2048$ . (iii) Size 6 (the middle coefficient). (iv) $2^{12} = 4096$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Middle term of $(a + b)^{10}$ ?Show answer
T₆.
2.Sum of binomial coefficients?Show answer
2ⁿ.
3.Odd-placed coefficient sum?Show answer
2ⁿ⁻¹.
4.Greatest coefficient of $(1 + x)^{8}$ ?Show answer
⁸C₄ = 70.

Middle Terms, Binomial Coefficients and Applications: frequently asked questions

Why does the middle term have the largest coefficient?

The binomial coefficients rise to the middle and then fall symmetrically.

How can I estimate a power like 1.02^6 quickly?

Use the first three terms of (1 + x)^n; the later terms are very small.

Where do divisibility results come from?

From writing the number as (1 + multiple of the divisor)ⁿ and noting that all terms beyond a few are multiples of the divisor.