Class 11 Maths · Original EZT practice
Binomial Theorem (Class 11 Chapter 7): summary and practice
The Binomial Theorem expands (a + b)^n without multiplying anything out: Pascal's triangle hands you the coefficients, the general term finds any one term directly, and the middle term lands at a fixed address. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Pascal's triangle
Each row adds the two entries above it, and row n holds exactly the coefficients of (a + b)^n: row 5 reads 1, 5, 10, 10, 5, 1. The triangle is the theorem's memory - no formula needed for small n.
2. The theorem itself
(a + b)^n = sum of C(n, k) a^(n-k) b^k for k from 0 to n. The powers of a fall while the powers of b rise, and the whole expansion carries n + 1 terms.
3. The general term
Term number k + 1 is T(k+1) = C(n, k) a^(n-k) b^k - the address of any single term, so you can pull out the x^2 coefficient or the constant term without writing the full expansion.
4. Middle terms
For even n there is one middle term, the (n/2 + 1)th; for odd n there are two, the ((n+1)/2)th and ((n+3)/2)th. In (a + b)^6 the middle is the 4th term, C(6, 3) a^3 b^3.
5. Special cases
Setting a = 1 and b = x gives (1 + x)^n with pure-C(n, k) coefficients; swapping b for -x alternates the signs. These two cover most quick problems, and the same machinery extends to (x + y)^n style sums.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Read a Pascal row
Choose the method: Row n lists C(n, 0) through C(n, n).
EZT practice question
Write row n = 5 and row n = 6 of Pascal's triangle.
Show the worked solution
- Start from 1, 4, 6, 4, 1 (n = 4).
- Add neighbours for the next row.
- Repeat once more.
Answer: n = 5: 1, 5, 10, 10, 5, 1. n = 6: 1, 6, 15, 20, 15, 6, 1.
Check: Each row is symmetric and its entries sum to 2^n - 32 and 64 here.
Watch for this: Row numbering starts at n = 0, a bare 1 - the sixth row of the triangle is the n = 5 row.
Now try it yourself
What is the second entry of row n = 7?
Check your answer
7.
The second entry of row n is always C(n, 1) = n.
Expand a binomial
Choose the method: Apply (a + b)^n with the coefficient row, tracking the inside coefficient.
EZT practice question
Expand (x + 2)^4.
Show the worked solution
- Row 4: 1, 4, 6, 4, 1.
- Terms: x^4, 4 x^3 (2), 6 x^2 (4), 4 x (8), 16.
- Multiply the 2-powers in.
Answer: x^4 + 8x^3 + 24x^2 + 32x + 16.
Check: Set x = 0 and the expansion must give 2^4 = 16 - it does.
Watch for this: Do not forget the inside number: the coefficient of x^3 is 4 x 2, not 4.
Now try it yourself
Expand (x - 1)^3.
Check your answer
x^3 - 3x^2 + 3x - 1.
Row 3 with alternating signs from the minus inside the bracket.
Pull out one coefficient
Choose the method: T(k+1) = C(n, k) a^(n-k) b^k; choose k to hit the power you want.
EZT practice question
Find the coefficient of x^2 in (1 + x)^5.
Show the worked solution
- The x^2 term is k = 2.
- Coefficient = C(5, 2).
- = (5 x 4) / 2.
Answer: 10.
Check: Pascal's n = 5 row reads 1, 5, 10, ... - the x^2 coefficient is the third entry, 10.
Watch for this: The term NUMBER is k + 1 - asking for the 3rd term and asking for k = 2 are the same request.
Now try it yourself
Coefficient of x^3 in (1 + x)^6?
Check your answer
20.
C(6, 3) = 20.
Locate the middle term
Choose the method: Even n: the (n/2 + 1)th term. Odd n: the two middle terms straddle the centre.
EZT practice question
Which term is the middle term of (a + b)^6, and what is it?
Show the worked solution
- n = 6 is even, so one middle term.
- Position: 6/2 + 1 = 4th term.
- T4 = C(6, 3) a^3 b^3.
Answer: The 4th term: 20 a^3 b^3.
Check: Seven terms in total, so the 4th sits dead centre with three on each side.
Watch for this: The middle term's POSITION is n/2 + 1, not n/2 - the count of terms is n + 1.
Now try it yourself
How many middle terms does (a + b)^5 have, and at which positions?
Check your answer
Two - the 3rd and 4th terms.
Odd n gives twin middles at (n+1)/2 = 3 and (n+3)/2 = 4.
Handle (a - b)^n
Choose the method: Replace b with -b; odd powers of b flip sign.
EZT practice question
What is the coefficient of the third term in (a - b)^4?
Show the worked solution
- Third term means k = 2.
- T3 = C(4, 2) a^2 (-b)^2.
- (-b)^2 = +b^2.
Answer: +6 - the term is 6 a^2 b^2.
Check: Signs in (a - b)^n alternate by k: +, -, +, -, + across the five terms.
Watch for this: Dropping the sign of -b - an odd power of b carries the minus into the term.
Now try it yourself
Sign of the 2nd term of (a - b)^4?
Check your answer
Negative: -4 a^3 b.
k = 1 is odd, so the minus survives: C(4, 1) a^3 (-b).
Count the terms
Choose the method: (a + b)^n has n + 1 terms after expansion.
EZT practice question
How many terms does the expansion of (a + b)^10 have, and what is the sum of the powers in each term?
Show the worked solution
- k runs from 0 to n.
- That is n + 1 terms.
- Each term holds a^(n-k) b^k, powers summing to n.
Answer: 11 terms, and the powers add to 10 in every one.
Check: The power count is the homogeneity check - if a term's powers do not sum to n, something slipped.
Watch for this: n + 1 terms, not n - counting from k = 0 is what adds the extra one.
Now try it yourself
Terms in (a + b)^7?
Check your answer
8.
n + 1 = 8.
Try a mixed set
Choose a method or explanation before opening each answer.
1. The theorem?
Show answer and reason
(a + b)^n = sum C(n, k) a^(n-k) b^k.
n + 1 terms total.
2. General term?
Show answer and reason
T(k+1) = C(n, k) a^(n-k) b^k.
Term number is k + 1.
3. Middle term, even n?
Show answer and reason
The (n/2 + 1)th term.
One centre for even n, two for odd.
4. Pascal row n = 5?
Show answer and reason
1, 5, 10, 10, 5, 1.
Each entry is the sum of the two above it.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find the coefficient of x^3 in (1 + x)^6.
- Expand (x - 1)^3.
- Write the middle term of (a + b)^6.
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Questions about this practice page
- How does Pascal's triangle give the coefficients?
- Row n of the triangle IS the list C(n, 0), C(n, 1), ..., C(n, n) - each entry is the sum of the two above it, which is exactly the identity C(n+1, k) = C(n, k-1) + C(n, k). For small n you can read the expansion straight off the row.
- Which term is the middle term?
- For even n there is one, at position n/2 + 1 - in (a + b)^6 that is the 4th term. For odd n the centre splits into two terms, at positions (n+1)/2 and (n+3)/2.
- Why does the general term use k + 1?
- Because k counts from 0 while term numbers count from 1. The first term carries k = 0, so term number k + 1 is the one holding b^k - keep the two straight or every answer shifts one place.
- Where is the binomial theorem actually used?
- Anywhere powers of sums appear: probability (binomial distributions), approximations like (1.01)^10 without a calculator, compound-interest style growth, and later the binomial series for negative and fractional indices.
- Are these the NCERT exercise questions for Binomial Theorem (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.