Chapter 7 of 14

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Odisha (BSE)Class 11 Mathematics★ Unit II · part of the 25-mark Algebra block shared with Complex Numbers and Quadratic Equations, Linear Inequalities, Permutations and Combinations, and Sequences and Series — CBSE states no chapter-wise split

Binomial Theorem

(98)^5 by hand? The Binomial Theorem writes it as (100-2)^5 and expands it in six lines using nCr coefficients, no repeated multiplication needed. This is the shortest chapter in the book — nine pages, one exercise, one miscellaneous exercise — and the topic most coaching material assumes it covers, finding a specific term via the general term formula, is not part of the current syllabus at all. CBSE Class 11 Mathematics Chapter 7.

⏱ 7 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Construct Pascal's triangle and state its entries as binomial coefficients nCr
2State the Binomial Theorem and reproduce its proof by mathematical induction, using Pascal's identity for the inductive step
3Expand (a+b)^n, (x-y)^n, (1+x)^n and (1-x)^n for a positive integer n
4Use the x=1 substitution to find the sum of binomial coefficients, and split that sum into its even-position and odd-position halves
5Apply the theorem to compute large powers exactly, approximate values using the first few terms, and prove divisibility-remainder statements

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Why this chapter matters

The Binomial Theorem turns an unwieldy repeated multiplication like (98)^5 into a six-term sum computed exactly by hand, and its proof by induction reuses Pascal's identity from the previous chapter directly. It is the shortest chapter in the book, but the general term and middle term formulas most coaching material treats as its centrepiece are formally formative-only here, not summative.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Combinations (nCr) and Pascal's identity nC(r-1)+nCr=(n+1)Cr from the Permutations and Combinations chapter, since the binomial coefficients and the induction proof both come directly from there

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Comfort expanding (a+b)^2 and (a+b)^3 by direct multiplication, since Pascal's triangle is built by generalising exactly that pattern

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The principle of mathematical induction — stating a base case and showing one case implies the next — used to prove the theorem itself

Binomial Theorem

1. Check this before you revise anything

The general term and middle term formulas — the tools coaching material leans on hardest for this chapter — are formally listed by CBSE as formative-only, not summative. The curriculum's dropped-topics block states plainly, under Binomial Theorem: "General and middle term in binomial expansion." The main course structure lists only: "Historical perspective, statement and proof of the binomial theorem for positive integral indices. Pascal's triangle, simple applications."

The current 2026-27 book matches this precisely. It has no section on finding a specific term, no "middle term" heading, and no second exercise. Its one and only exercise is Exercise 7.1, followed by a six-question Miscellaneous Exercise.

One thing this chapter does better than several of its neighbours: it doesn't just state the theorem, it proves it. Section 7.2 walks through a full induction proof, base case through inductive step, and that proof leans on Pascal's identity ( $^{k} C_{r} +^{k} C_{r - 1} =^{(k + 1)} C_{r}$ ) — the exact identity the Permutations and Combinations chapter derived. If a "prove the binomial theorem" question shows up, that's genuinely summative content, not an overreach.

2. What this chapter covers

Textbook section	Topic
7.1	Introduction — why repeated multiplication breaks down for large powers
7.2	Binomial Theorem for positive integral indices: statement, induction proof, Pascal's triangle, special cases
Exercise 7.1	Expanding binomials; evaluating large powers via the theorem; proof-style questions
Miscellaneous Exercise	Mixed expansion, approximation, and divisibility problems

3. Pascal's triangle

Writing out $(a + b)^{0}$ through $(a + b)^{4}$ by direct multiplication and lining up their coefficients produces a triangular array, where every interior entry is the sum of the two entries directly above it:

$1$ $11$ $121$ $1331$ $14641$

Row $n$ (counting the top row as row 0) gives the coefficients of $(a + b)^{n}$ . Once combinations are available, every entry in row $n$ is just $^{n} C_{r}$ for $r = 0, 1, \dots, n$ — row 7, for instance, is $^{7} C_{0},^{7} C_{1}, \dots,^{7} C_{7}$ , with no need to write out rows 0 through 6 first to get there.

Worked, mirroring the textbook's own construction. To expand $(2 x + 3 y)^{5}$ : row 5 of Pascal's triangle is $1, 5, 10, 10, 5, 1$ . Applying these as coefficients, with the power of $2 x$ falling from 5 to 0 while the power of $3 y$ rises from 0 to 5: $(2 x + 3 y)^{5} = 32 x^{5} + 240 x^{4} y + 720 x^{3} y^{2} + 1080 x^{2} y^{3} + 810 x y^{4} + 243 y^{5}$ .

4. The Binomial Theorem

For any positive integer $n$ :

$(a + b)^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + \dots +^{n} C_{n - 1} a b^{n - 1} +^{n} C_{n} b^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$

Proof sketch (by the principle of mathematical induction), matching the book's own argument. For $n = 1$ : $(a + b)^{1} =^{1} C_{0} a +^{1} C_{1} b = a + b$ , true. Assume the statement holds for some $n = k$ . Multiplying both sides by $(a + b)$ and grouping like powers, each new coefficient becomes a sum of two consecutive coefficients from the $k$ -th row: $^{k} C_{r} +^{k} C_{r - 1}$ .

By Pascal's identity, this equals $^{(k + 1)} C_{r}$ exactly, which is what the $(k + 1)$ -th row requires. Since the base case holds and each case implies the next, the theorem holds for every positive integer $n$ .

Five observations worth holding onto, all direct consequences of the formula:

The expansion has exactly $n + 1$ terms.
The power of $a$ falls from $n$ to $0$ ; the power of $b$ rises from $0$ to $n$ ; every term's two exponents add to $n$ .
The coefficients $^{n} C_{r}$ are called binomial coefficients, and $^{n} C_{0} =^{n} C_{n} = 1$ always bookends the row.

Worked, mirroring the textbook's own illustration. Expanding $(x + 2)^{6}$ : $^{6} C_{0} x^{6} +^{6} C_{1} x^{5} (2) +^{6} C_{2} x^{4} (2^{2}) +^{6} C_{3} x^{3} (2^{3}) +^{6} C_{4} x^{2} (2^{4}) +^{6} C_{5} x (2^{5}) +^{6} C_{6} (2^{6}) = x^{6} + 12 x^{5} + 60 x^{4} + 160 x^{3} + 240 x^{2} + 192 x + 64$ .

5. Special cases

Taking $b = - y$ gives an alternating-sign version: $(x - y)^{n} =^{n} C_{0} x^{n} -^{n} C_{1} x^{n - 1} y +^{n} C_{2} x^{n - 2} y^{2} - \dots + (- 1)^{n}^{n} C_{n} y^{n}$ .

Taking $a = 1, b = x$ gives: $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n}$ . Substituting $x = 1$ : $2^{n} =^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n}$ — the sum of every binomial coefficient in row $n$ .

Taking $a = 1, b = - x$ gives: $(1 - x)^{n} =^{n} C_{0} -^{n} C_{1} x +^{n} C_{2} x^{2} - \dots + (- 1)^{n}^{n} C_{n} x^{n}$ . Substituting $x = 1$ : $0 =^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots + (- 1)^{n}^{n} C_{n}$ — the alternating sum vanishes.

A corollary the book sets up but doesn't spell out by name: adding and subtracting those last two identities separates row $n$ into its even- and odd-position coefficients. Adding them cancels every odd-position term, leaving $2 \times (sum of even-position coefficients) = 2^{n}$ ; subtracting cancels every even-position term instead. Either way, sum of even-position coefficients = sum of odd-position coefficients = $2^{n - 1}$ .

6. Worked applications

Mirroring the textbook's own Example 1. Expand $(x^{2} + \frac{3}{x})^{4}$ , $x \neq = 0$ : $=^{4} C_{0} (x^{2})^{4} +^{4} C_{1} (x^{2})^{3} (\frac{3}{x}) +^{4} C_{2} (x^{2})^{2} (\frac{3}{x})^{2} +^{4} C_{3} (x^{2}) (\frac{3}{x})^{3} +^{4} C_{4} (\frac{3}{x})^{4} = x^{8} + 12 x^{5} + 54 x^{2} + \frac{108}{x} + \frac{81}{x ^{4}}$ .

Mirroring the textbook's own Example 2 — using the theorem to compute a large power exactly, no calculator needed. Compute $(98)^{5}$ by writing $98 = 100 - 2$ : $(100 - 2)^{5} =^{5} C_{0} (100)^{5} -^{5} C_{1} (100)^{4} (2) +^{5} C_{2} (100)^{3} (2)^{2} -^{5} C_{3} (100)^{2} (2)^{3} +^{5} C_{4} (100) (2)^{4} -^{5} C_{5} (2)^{5} = 10040008000 - 1000800032 = 9039207968$ .

Mirroring the textbook's own Example 4 — a divisibility proof built entirely from the $(1 + x)^{n}$ special case. Prove $6^{n} - 5^{n}$ always leaves remainder 1 on division by 25. Write $6^{n} = (1 + 5)^{n} =^{n} C_{0} +^{n} C_{1} (5) +^{n} C_{2} (5^{2}) + \dots +^{n} C_{n} (5^{n})$ , so $6^{n} - 5 n = 1 + 5^{2} (^{n} C_{2} +^{n} C_{3} (5) + \dots + 5^{n - 2}) = 25 k + 1$ for some natural number $k$ . Since $6^{n} - 5^{n} = 25 k + 1$ , dividing by 25 always leaves remainder 1.

This last technique — isolate the first two terms of $(1 + x)^{n}$ explicitly and bound or discard the rest — is the same move behind comparing $(1.01)^{1, 000, 000}$ against 10,000: since every term beyond the first two is positive, $(1 + 0.01)^{1, 000, 000} > 1 + 1, 000, 000 (0.01) = 10, 001 > 10, 000$ , without ever expanding the full sum.

Summary

$(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ , proved by induction using Pascal's identity $^{k} C_{r} +^{k} C_{r - 1} =^{(k + 1)} C_{r}$ ; the expansion always has $n + 1$ terms.
Pascal's triangle is the same coefficient array read a different way: row $n$ , entry $r$ , is $^{n} C_{r}$ .
$(x - y)^{n}$ alternates sign; $(1 + x)^{n}$ at $x = 1$ gives $2^{n} =$ sum of all coefficients; $(1 - x)^{n}$ at $x = 1$ gives $0 =$ the alternating sum — combined, these show the even-position and odd-position coefficients each sum to $2^{n - 1}$ .
Writing a number as a sum or difference of two easy-to-power values (like $98 = 100 - 2$ ) turns an unwieldy computation into a short binomial expansion.
Isolating the first two terms of $(1 + x)^{n}$ , with the rest bounded or discarded, is the standard technique for large-power comparisons and divisibility-remainder proofs.
Finding a specific term via the general term formula, or a middle term, is formative-only in the current syllabus — it is not covered in this chapter's own content or exercises.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Binomial Theorem

(a+b)^n = nC0.a^n + nC1.a^(n-1)b + nC2.a^(n-2)b^2 + ... + nCn.b^n = sum_(r=0)^n nCr.a^(n-r).b^r

Has exactly n+1 terms; proved by induction, with the inductive step using Pascal's identity nCr+nC(r-1)=(n+1)Cr

Pascal's triangle row-coefficient link

Row n, entry r, of Pascal's triangle = nCr

Lets any row be written directly from combinations, without building every earlier row first

Alternating-sign special case

(x-y)^n = nC0.x^n - nC1.x^(n-1)y + nC2.x^(n-2)y^2 - ... + (-1)^n.nCn.y^n

Obtained by substituting b=-y in the main theorem; signs alternate starting positive

(1+x)^n special case

(1+x)^n = nC0 + nC1.x + nC2.x^2 + ... + nCn.x^n

At x=1: 2^n = nC0+nC1+...+nCn, the sum of every binomial coefficient in row n

(1-x)^n special case

(1-x)^n = nC0 - nC1.x + nC2.x^2 - ... + (-1)^n.nCn.x^n

At x=1: 0 = nC0-nC1+nC2-..., the alternating sum of row n's coefficients

Even/odd coefficient split

nC0+nC2+nC4+... = nC1+nC3+nC5+... = 2^(n-1)

Add and subtract the two special-case identities above at x=1 to cancel the odd or even terms respectively

Exact large-power computation

Write the base as a sum/difference of two easy powers (e.g. 98=100-2), then expand with the theorem

Every term is exact — no rounding — since the theorem is an identity, not an approximation

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Expecting a 'find the 5th term' or 'find the middle term' question from this chapter

✓ The general term and middle term formulas are formative-only in the current syllabus — this chapter's own content and its one exercise never ask for a specific term. That skill belongs to material outside this book, not to CBSE's summative assessment of it.

WATCH OUT

✗ Dropping the alternating signs when expanding (x-y)^n

✓ Every odd-r term is negative: nC0.x^n - nC1.x^(n-1)y + nC2.x^(n-2)y^2 - ... Write out the sign row (+,-,+,-,...) before substituting values, rather than trying to track it mentally.

WATCH OUT

✗ Substituting x=1 into (1+x)^n but forgetting this only gives the sum of coefficients, not the value of the expansion at some other point the question actually asked for

✓ The x=1 substitution is specifically for 'sum of coefficients' questions. If the question gives a different value of x, substitute that value instead — 2^n only applies at x=1.

WATCH OUT

✗ Losing track of signs across many terms when writing a number as a difference and expanding, e.g. 98=100-2

✓ List every term's sign explicitly before computing any arithmetic — with 5 or 6 terms and alternating signs, mental sign-tracking is where most errors creep in, not the multiplication itself.

WATCH OUT

✗ Including too many or too few terms in an 'approximate using the first three terms' question

✓ Count exactly as many terms as the question specifies, starting from nC0, and state clearly that the result is an approximation, not the exact value — the terms left out are what makes it approximate.

WATCH OUT

✗ Stopping a divisibility proof once the expression is written as 25k+1 (or similar) without stating what that means

✓ The algebra alone doesn't answer the question — finish with a sentence: 'since 6^n-5n=25k+1, dividing by 25 always leaves remainder 1,' connecting the algebraic form back to the word 'remainder'.

WATCH OUT

✗ Treating the induction proof of the theorem as background material rather than examinable content

✓ The syllabus explicitly lists 'statement and proof of the binomial theorem' — unlike the theorem's applications, the proof itself (base case, inductive step using Pascal's identity) can be asked for directly.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 7.1

Exercise 7.1

Expanding binomial expressions using the theorem, evaluating large powers exactly, and proof-style questions built from its special cases

Questions

Miscellaneous Ex

Miscellaneous Exercise

Divisibility proofs, symmetric-expression evaluation, approximation using the first few terms, and expansion of a trinomial by grouping it into a binomial

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Binomial Theorem?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min worth ~25 marks in Odisha (BSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•(a+b)^n = sum of nCr.a^(n-r).b^r for r=0 to n, proved by induction using Pascal's identity nCr+nC(r-1)=(n+1)Cr
•Pascal's triangle row n, entry r, equals nCr — the same coefficients, a different way to read them off
•(x-y)^n alternates sign; (1+x)^n at x=1 gives 2^n = sum of all coefficients; (1-x)^n at x=1 gives 0 = the alternating sum
•Adding/subtracting those last two identities splits row n into even-position and odd-position halves, each summing to 2^(n-1)
•Writing a number as a sum or difference of two easy-to-power values turns an unwieldy computation into an exact short expansion
•Isolating the first two (or three) terms of (1+x)^n, with the rest bounded or discarded, drives large-power comparisons, approximations, and divisibility-remainder proofs alike
•The general term and middle term formulas are formative-only in the current syllabus — not covered in this chapter's content or its one exercise

Odisha (BSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Part of Unit II's 25-mark Algebra block (no chapter-wise split, per CBSE)

Question type	Marks each	Typical count	What it tests
Pascal's Triangle and Binomial Expansion	2-3	1	Constructing a row of Pascal's triangle, direct expansion of (a+b)^n for a small positive integer n
Evaluating Powers Using the Binomial Theorem	3-4	1	Writing a number as a sum/difference of two easy powers and expanding exactly, e.g. (102)^4 or (98)^5
Special Cases and Coefficient Sum Applications	3-5	1	Sum of binomial coefficients at x=1, even/odd coefficient split, symmetric surd-expression evaluation via (a+b)^n +/- (a-b)^n
Divisibility and Approximation Proofs	4-6	0-1	Proving a remainder statement from the (1+x)^n expansion, or approximating a value using a stated number of terms

Prep strategy

Practise writing out Pascal's triangle rows quickly up to row 8 or so — it's the fastest sanity check on any small-n expansion
For every special-case question, identify which substitution (x=1 in (1+x)^n, or a=1,b=-x, etc.) the question is actually asking for before writing anything
For divisibility proofs, always finish with an explicit sentence connecting the algebraic form back to the word 'remainder' or 'divisible'

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Probability distributions

The binomial probability formula P(X=r) = nCr.p^r.(1-p)^(n-r) for r successes in n independent trials is a direct reuse of the binomial coefficients this chapter builds.

Compound interest and small-percentage growth

Approximating (1+r)^n using the first two or three terms of its expansion is exactly how small compound-growth rates are estimated by hand before a calculator is reached for.

Error and tolerance estimates in measurement

When a measured quantity has a small relative error x, expressions like (1+x)^n are approximated using the first couple of binomial terms to estimate how that error propagates through a computed power.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

For 'expand' questions, write the relevant Pascal's triangle row first — it turns the rest of the problem into substitution, not memorisation

For 'evaluate using the binomial theorem' questions, split the number into a sum/difference of two round values before doing anything else

For any x=1 substitution question, state explicitly which identity is being used (sum of coefficients vs the alternating-sum corollary) before computing

Never introduce a general-term or middle-term formula in an answer — it isn't part of this chapter's syllabus and won't match the mark scheme

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Bounding (1+1/n)^n between 2 and 3 using only the expansion (no limits or calculus) is the classical elementary proof that this sequence is bounded, a first step toward the constant e

STRETCH

The multinomial extension (a+b+c)^n = sum of n!/(p!q!r!).a^p.b^q.c^r for p+q+r=n generalises the same induction idea to three or more terms

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainDivisibility proof from the (1+x)^n special caseProof by direct expansion

Using the Binomial Theorem, prove that $1 1^{n} - 10 n - 1$ is divisible by 100 for every positive integer $n$ .

Stuck? Show the approach

Write $1 1^{n} = (1 + 10)^{n}$ , expand using the theorem, and isolate the first two terms — the rest of the sum will carry a common factor of $100$ .

Show the full solution

$(1 + 10)^{n} =^{n} C_{0} +^{n} C_{1} (10) +^{n} C_{2} (10)^{2} + \dots +^{n} C_{n} (10)^{n} = 1 + 10 n + 100 (^{n} C_{2} + 10 \cdot^{n} C_{3} + \dots + 1 0^{n - 2} \cdot^{n} C_{n})$ . So $1 1^{n} - 10 n - 1 = 100 (^{n} C_{2} + 10 \cdot^{n} C_{3} + \dots)$ , a multiple of 100.

Answer: Proved: 11^n - 10n - 1 = 100k for a natural number k

The trap

Stopping once the expression is written as 100 times something, without stating explicitly that this proves divisibility by 100, leaves the proof looking incomplete even when the algebra is correct.

JEE MainFinding n from the sum of binomial coefficientsReverse application of a special case

If the sum of the binomial coefficients in the expansion of $(x + y)^{n}$ is 256, find $n$ , and state how many terms the expansion has.

Stuck? Show the approach

The sum of all binomial coefficients in row $n$ is $2^{n}$ , obtained by setting the variable ratio to 1 — solve $2^{n} = 256$ for $n$ , then use $n + 1$ for the term count.

Show the full solution

$2^{n} = 256 = 2^{8}$ , so $n = 8$ . The expansion of $(x + y)^{8}$ has $n + 1 = 9$ terms.

Answer: n = 8, with 9 terms in the expansion

The trap

Confusing 'sum of coefficients' with 'sum of terms' — the sum of coefficients comes from the x=1 substitution and is a single number (256 here), not related to how many terms the expansion has, which is a separate count (n+1).

JEE MainEvaluating a symmetric surd expressionEven-term survival identity

Find the value of $(5 + 2)^{6} + (5 - 2)^{6}$ .

Stuck? Show the approach

Adding $(a + b)^{n}$ and $(a - b)^{n}$ cancels every odd-position term and doubles every even-position term, so only $r = 0, 2, 4, 6$ need to be computed.

Show the full solution

With $a = 5, b = 2, n = 6$ : $r = 0$ : $^{6} C_{0} a^{6} = 125$ . $r = 2$ : $^{6} C_{2} a^{4} b^{2} = 15 (25) (2) = 750$ . $r = 4$ : $^{6} C_{4} a^{2} b^{4} = 15 (5) (4) = 300$ . $r = 6$ : $^{6} C_{6} b^{6} = 8$ . Sum $= 125 + 750 + 300 + 8 = 1183$ ; doubling gives $2366$ .

Answer: 2366

The trap

Forgetting to double the bracketed sum (the identity gives twice the even-position sum, not the sum itself) is the most common arithmetic slip in this technique.

JEE MainApproximating a large power to a stated number of termsTruncated expansion

Using the first three terms of its binomial expansion, find the approximate value of $(1.02)^{8}$ .

Stuck? Show the approach

Write $1.02 = 1 + 0.02$ and expand with the theorem, keeping only the first three terms ( $r = 0, 1, 2$ ) since later terms involve higher powers of a small number and contribute far less.

Show the full solution

$(1 + 0.02)^{8} \approx^{8} C_{0} +^{8} C_{1} (0.02) +^{8} C_{2} (0.02)^{2} = 1 + 8 (0.02) + 28 (0.0004) = 1 + 0.16 + 0.0112 = 1.1712$ .

Answer: approximately 1.1712

The trap

Including a fourth term 'to be safe' changes the answer the question is actually asking for — 'first three terms' is a precise instruction, not a suggested minimum.

JEE AdvancedBounding a binomial expansion to prove an inequalityMulti-step bounding proof

Prove that $(1 + \frac{1}{n})^{n} < 3$ for every positive integer $n$ .

Stuck? Show the approach

Expand using the theorem, rewrite each term's product of falling factors as a product of factors each less than 1, bound every term above by $1/ r!$ , then bound the resulting factorial sum by a geometric series.

Show the full solution

$(1 + \frac{1}{n})^{n} = r = 0 \sum n^{n} C_{r} \frac{1}{n ^{r}} = r = 0 \sum n \frac{1}{r !} \cdot each factor < 1 for r \geq 1 \frac{n ( n - 1 ) \dots ( n - r + 1 )}{n ^{r}}$ . So every term is less than $1/ r!$ , giving $(1 + \frac{1}{n})^{n} < r = 0 \sum n \frac{1}{r !} = 1 + 1 + \frac{1}{2 !} + \frac{1}{3 !} + \dots$ . Since $r! \geq 2^{r - 1}$ for $r \geq 1$ , this sum is at most $1 + (1 + \frac{1}{2} + \frac{1}{4} + \dots) = 1 + 2 = 3$ .

Answer: Proved: (1+1/n)^n < 3 for every positive integer n

The trap

Trying to prove this by testing individual values of n only ever confirms it for those specific cases — the factor-by-factor bound on each term is what makes the argument work for every n at once.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardMedium

JEE MainHigh

JEE AdvancedMedium

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No. It's listed by CBSE as formative-only, and the current 2026-27 book has no section on it and no exercise question asking for a specific term. It's a genuinely useful tool for competitive exams, but it isn't part of this chapter's summative CBSE content.

Because the syllabus explicitly names both: 'statement and proof of the binomial theorem.' The proof itself, by induction using Pascal's identity, is examinable content, not optional background reading.

Substitute x=1 into (1+x)^n = nC0+nC1.x+...+nCn.x^n. Every coefficient becomes multiplied by 1, so the sum is just (1+1)^n = 2^n.

Write 99 as 100-1, expand (100-1)^5 with the theorem, and compute each term exactly — every term after the first two or three shrinks fast since they involve higher powers of the small number, but none are dropped since this is an exact computation, not an approximation.

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Every later chapter in this book — functions, trigonometry, probability, even calculus — is written in a language, and that language is set notation. This chapter builds that language from a single idea (a well-defined collection of objects) up through subsets, Venn diagrams, and the operations that combine sets, ending with De Morgan's laws. One formula that shows up in almost every coaching handout for this chapter is not actually in the 2026-27 book at all. CBSE Class 11 Mathematics Chapter 1.

Relations and Functions

A relation is any association between two sets at all — as loose or as strange as you like. A function is the one, very specific kind of relation where every input has exactly one output, and that single restriction is what makes it possible to build the rest of mathematics on top of it. One classification almost every coaching handout drills for this chapter — one-one, onto, bijective — has actually moved to Class 12. CBSE Class 11 Mathematics Chapter 2.

Trigonometric Functions

The unit-circle definition extends sine and cosine, first met as ratios inside a right triangle, to every real number — including negative numbers and numbers far larger than 360 degrees. One topic many coaching sheets drill hard for this chapter, the general solution of trigonometric equations, is graded formatively only under the 2026-27 CBSE syllabus and never appears in this chapter's own exercises. CBSE Class 11 Mathematics Chapter 3.

Complex Numbers and Quadratic Equations

The equation x² + 1 = 0 has no real solution, so this chapter builds a new number, i = √(-1), and a new algebra around it. Despite the chapter's own name, actually solving ax² + bx + c = 0 for complex roots — and the polar form its own section heading promises — are both graded formatively only under the 2026-27 CBSE syllabus. CBSE Class 11 Mathematics Chapter 4.

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