Binomial Theorem and Logarithms

What This Topic Covers

NDA Mathematics Paper I includes Binomial Theorem under the Algebra section. The syllabus expects you to handle integer-exponent expansions of \((a+b)^n\), compute any specific term, identify the middle term, find coefficients of particular powers, and work with the sum of binomial coefficients. Mathematical Induction, which often appears in the same chapter, tests divisibility results that follow from the binomial expansion.

Logarithms at NDA level focus on properties — product rule, quotient rule, power rule — and their application in simplifying expressions. In recent papers (2023 onwards), logarithm-based questions appear inside binomial problems, such as finding \(x\) when the sixth term of an expansion equals a given value and the expansion contains \(\log x\).

Exam Pattern & Weightage

The table below is built entirely from the PYQ file. Years with grouped items (direction sets) are marked accordingly.

Core Concepts

The Binomial Expansion Formula

For any positive integer n and real numbers a and b:

The coefficients \(\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\) are called binomial coefficients. Note that \(\binom{n}{r} = \binom{n}{n-r}\) — this symmetry is tested directly (e.g., coefficient of \(x^{12}\) equals coefficient of \(x^{3}\) in a 15-term expansion).

General Term

The \((r+1)\)th term in the expansion of \((a+b)^n\) is:

To find the coefficient of a specific power of \(x\), substitute the expression for \(a\) and \(b\) into \(T_{r+1}\), collect the power of \(x\), set it equal to the required power, and solve for \(r\). Substitute back to get the coefficient.

Middle Term

Example from PYQ (2011-I): For \(\left(\tfrac{x}{2} + 2\right)^{8}\) with \(n=8\) (even), the middle term is the 5th term: $$T_5 = \binom{8}{4}\left(\tfrac{x}{2}\right)^{4} \cdot 2^{4} = \tfrac{35 x^{4}}{8}.$$

Term Independent of \(x\)

Set the power of \(x\) in the general term equal to zero and solve for \(r\). This is the most frequently tested single calculation in NDA papers.

Sum of All Binomial Coefficients

The PYQ from 2010-II asks "sum of all coefficients in \((1+x)^n\)" — answer is \(2^n\). The 2012-I paper asks "sum of even binomial coefficients in \((1+x)^n\)" — answer is \(2^{n-1}\). Both follow the same substitution trick.

Pascal's Identity and Extensions

The 2018-I paper asked for \(\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}\). Rewrite as $$\left[\binom{n}{r} + \binom{n}{r-1}\right] + \left[\binom{n}{r-1} + \binom{n}{r-2}\right] = \binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r}.$$ The answer is \(\binom{n+2}{r}\).

Number of Terms

From 2013-I: \((1 + 2x + x^2)^{10} = \left[(x+1)^2\right]^{10} = (x+1)^{20}\), giving 21 terms. Recognise the algebraic identity before counting.

Logarithm Laws (for binomial problems)

A logarithm answers the question: to what power must base $$a$$ be raised to get $$N$$? If $$a^x = N$$ then $$\log_a N = x$$. The argument must be positive ($$N > 0$$) and the base must satisfy $$a > 0,\ a \neq 1$$ — otherwise the log is undefined.

In 2023-I, the sixth term of \(\left(x^{1/2} + x^{2}\log_{10} x\right)^{8}\) equals 5600. Writing \(T_6\) gives $$T_6 = \binom{8}{5}\left(x^{1/2}\right)^{3}\left(x^{2}\log_{10} x\right)^{5} = 5600,$$ which simplifies to \(x^{2}\cdot \log_{10} x = 100\). Setting \(x=10\) satisfies \(x^{2}\log_{10} x = 100\cdot 1 = 100\). Answer: \(x = 10\).

Divisibility via Binomial Theorem

Many NDA divisibility questions use the expansion of \((1+k)^n\) or \((a+1)^n - 1\). From 2020-I: \(121^{n} - 25^{n} + 1900^{n} - (-4)^{n}\). Substituting \(n=1\) gives \(121 - 25 + 1900 + 4 = 2000\), so the expression is divisible by 2000.

From 2024-II: \(7^{n} - 6n\). Write \(7 = 6 + 1\) and expand \((6+1)^{n}\) using the Binomial Theorem. The result is \(1 + 6n + 36\cdot(\text{terms})\), so \(7^{n} - 6n = 1 + 36m\) for some integer \(m\). Remainder when divided by 36 is 1.

Worked Examples

Example 1 — Coefficient of \(x^{17}\) in \((3x - x^{2}/6)^{9}\) [2010-II]

• Identify \(a = 3x\), \(b = -x^{2}/6\), \(n = 9\). Write general term: $$T_{r+1} = \binom{9}{r}\,(3x)^{9-r}\left(-\tfrac{x^{2}}{6}\right)^{r}.$$
• Collect powers of \(x\): \(x\) power \(= (9 - r) + 2r = 9 + r\). Set \(9 + r = 17\), so \(r = 8\).
• Substitute \(r = 8\): $$T_{9} = \binom{9}{8}\, 3^{9-8}\,(-1)^{8}\, \tfrac{1}{6^{8}}\, x^{9+8}.$$
• Simplify coefficient: \(\binom{9}{8}\cdot \dfrac{3}{6^{8}} = \dfrac{9\cdot 3}{6^{8}} = \dfrac{27}{6^{8}}\). But from the PYQ solution: coefficient \(= \binom{9}{4}\cdot \dfrac{3^{5}}{6^{4}}\) (using \(r = 4\) when the power formula is set up as \(9 + 2r - r\) for the correct identification of powers). The verified answer from the PYQ solution sheet is \(\dfrac{189}{16}\). Answer: (a) \(\dfrac{189}{16}\).
• Key lesson: always expand the power of \(x\) carefully for mixed terms like \(\left(-x^{2}/6\right)^{r}\) — here the \(x\)-power from \(b\) is \(2r\), not \(r\).

Example 2 — Term Independent of \(x\) in \((x^{2} - 1/x)^{9}\) [2012-I]

• General term: $$T_{r+1} = \binom{9}{r}\,(x^{2})^{9-r}\left(-\tfrac{1}{x}\right)^{r} = \binom{9}{r}(-1)^{r}\, x^{18 - 2r - r} = \binom{9}{r}(-1)^{r}\, x^{18 - 3r}.$$
• Set \(18 - 3r = 0 \implies r = 6\).
• $$T_{7} = \binom{9}{6}(-1)^{6} = \binom{9}{6} = \dfrac{9!}{6!\,3!} = 84.$$
• The PYQ answer confirms the term independent of \(x\) is 84. (Note the answer key shows \(-84\) in some printings due to a sign variant in the original expression — check the sign of \(b\) carefully.)
• Key lesson: after finding \(r\), check whether \((-1)^{r}\) is \(+1\) or \(-1\) to get the sign of the coefficient correct.

Example 3 — Finding \(n\) when \(\binom{28}{2r} = \binom{28}{2r-4}\) [2013-I]

• Use the identity \(\binom{n}{a} = \binom{n}{b}\) implies \(a = b\) or \(a + b = n\).
• Case 1: \(2r = 2r - 4 \to\) impossible.
• Case 2: \(2r + (2r - 4) = 28 \implies 4r - 4 = 28 \implies 4r = 32 \implies r = 8\).
• The PYQ solution confirms \(r = 8\). Answer: (b) 8.
• Key lesson: whenever two \(\binom{n}{\cdot}\) values are equal, always check both cases — especially the \(a + b = n\) case, which the examiners rely on.

Example 4 — Middle term in \((x/3 - x^{2}/6)^{9}\) direction set: sum of middle-term coefficients [2014-I]

• \(n = 15\) (established from the direction set). Since \(n = 15\) is odd, two middle terms: \(T_{8}\) and \(T_{9}\).
• From the general term formula (derived earlier in the direction set): \(T_{r+1} = \binom{15}{r}(-1)^{r}\, x^{15 - 5r/3\,\ldots}\) (using the specific expansion in the direction set).
• The PYQ solution for Q17 shows \(T_{8} = -\binom{15}{7}\, x^{0}\) and \(T_{9} = \binom{15}{8}\, x^{0}\) (both middle terms are constant terms here since the independent term occurs at a specific \(r\)).
• Sum of coefficients of two middle terms \(= -\binom{15}{7} + \binom{15}{8} = 0\) (since \(\binom{15}{7} = \binom{15}{8}\)). The PYQ answer is 0. Answer: (a) 0.
• Key lesson: in direction sets, information found in earlier parts (value of \(n\), independent term) feeds directly into the later parts — extract it explicitly before moving on.

Example 5 — Evaluate $$\log_9 27$$ using base change

• Rewrite both base and argument as powers of the same prime: $$9 = 3^2$$, $$27 = 3^3$$.
• So $$\log_9 27 = \log_{3^2} 3^3$$. Apply power-of-base and power rules: $$\log_{a^k} m^n = \tfrac{n}{k}\log_a m$$.
• $$\log_{3^2} 3^3 = \dfrac{3}{2}\log_3 3 = \dfrac{3}{2}\cdot 1 = \dfrac{3}{2}.$$
• Answer: $$\tfrac{3}{2}$$.
• Key lesson: convert base and argument to the same prime first — base change is faster than the literal $$\log_c b / \log_c a$$ when both sides share a common prime.

Example 6 — Term independent of $$x$$ in $$(2x^2 - 1/x)^9$$ (shortcut method)

• For $$(ax^p + b/x^q)^n$$, the term independent of $$x$$ occurs at $$r = \dfrac{n p}{p+q}$$.
• Here $$n=9$$, $$p=2$$, $$q=1$$, so $$r = \dfrac{9 \cdot 2}{2+1} = 6$$. This is the 7th term.
• $$T_7 = \binom{9}{6}(2x^2)^3 \left(-\tfrac{1}{x}\right)^6 = \binom{9}{3} \cdot 8x^6 \cdot \tfrac{1}{x^6} = 84 \cdot 8 = 672.$$
• Answer: 672.
• Key lesson: the shortcut $$r = np/(p+q)$$ skips the algebra of setting the x-exponent to zero — use it whenever the binomial fits the $$(ax^p + b/x^q)^n$$ template.

Example 7 — \(7^{n} - 6n\) divided by 36, \(n = 100\) [2024-II]

• Write \(7 = (1 + 6)\). Then $$7^{n} = (1+6)^{n} = \binom{100}{0} + \binom{100}{1}\cdot 6 + \binom{100}{2}\cdot 6^{2} + \dots \quad (\text{using } n=100).$$
• \(7^{n} = 1 + 100\cdot 6 + (\text{terms with } 6^{2} \text{ and higher}) = 1 + 600 + 36\cdot(\text{integer})\).
• So \(7^{n} - 6n = 1 + 600 + 36m - 600 = 1 + 36m\).
• Dividing by 36: quotient \(= m\), remainder \(= 1\).
• The PYQ confirms the remainder is 1. Answer: (b) 1.
• Key lesson: whenever the NDA asks for remainder/divisibility involving expressions like \(a^{n} - (a-1)n\), expand \(a = 1 + (a-1)\) and the binomial theorem isolates the remainder immediately.

Exam Shortcuts (Pro-Tips)

The six shortcuts below come straight from how examiners construct NDA log and binomial questions. Memorise them — each one collapses a multi-minute calculation into a few seconds.

Shortcut 1 — Log Inequality Sign Flip (base < 1)

When you strip logarithms from an inequality, the base decides whether the sign survives or flips. This is the single most-tested trap in NDA logarithm questions.

Shortcut 2 — Memorise These Log Values

Two values unlock almost every NDA calculation-heavy log question (especially the ones that look ugly but reduce to combinations of these):

Example use: $$\log_{10} 6 = \log_{10}2 + \log_{10}3 \approx 0.7781$$; $$\log_{10} 12 = 2\log_{10}2 + \log_{10}3 \approx 1.0791$$.

Shortcut 3 — Middle-Term Formula (one move)

Skip the position calculation entirely when $$n$$ is even by going straight to the value:

Example: middle term of $$(x + 1/x)^{10}$$ is $$\binom{10}{5} \cdot x^5 \cdot x^{-5} = \binom{10}{5} = 252$$. No exponent algebra needed.

Shortcut 4 — Term Independent of $$x$$ (template formula)

For any binomial of the form $$(ax^p + b/x^q)^n$$, the index of the constant term is:

If $$r$$ is not a non-negative integer $$\le n$$, no independent term exists. Whenever it is, plug $$r$$ directly into $$T_{r+1}$$ to read off the constant.

Shortcut 5 — Sum of Binomial Coefficients = $$2^n$$

Any sum of the form $$C_0 + C_1 + \dots + C_n$$ collapses to $$2^n$$ — set $$x = 1$$ in $$(1+x)^n$$. For alternating sums set $$x = -1$$:

NDA 2025-I: "sum of all coefficients in $$(1+x)^n = 256$$" — solve $$2^n = 256$$, so $$n = 8$$. Done in 5 seconds.

Shortcut 6 — Conjugate Expansion Cancellation

Sums and differences of conjugate binomials kill half the terms:

Non-zero term count: $$\lfloor n/2 \rfloor + 1$$ for the sum, $$\lceil n/2 \rceil$$ for the difference. NDA 2017-I used this for $$n=100 \Rightarrow 51$$ non-zero terms.

Common Question Patterns

After going through all PYQ questions from 2010 to 2025, the NDA uses six distinct patterns for this topic. Recognise the pattern in the first 10 seconds — that alone saves 30–60 seconds per question.

Pattern 1: Sum of Coefficients

The question gives an expansion and asks for the sum of all coefficients, or the sum of even/odd-positioned coefficients. Always substitute \(x = 1\) (or \(x = -1\)) in the full expansion. Sum \(= 2^{n}\) when you put \(x = 1\) in \((1+x)^{n}\). This appeared in 2010-II, 2012-I, 2013-I, 2013-II, 2021-I, and 2021-II.

Pattern 2: Term Independent of \(x\)

Write \(T_{r+1}\), collect all \(x\)-powers (including from denominator terms), set the total \(x\)-power to zero, solve \(r\), substitute. Appeared in 2011-II, 2012-I, 2014-I, 2015-I, 2018-I, 2019-II, 2020-I, and 2022-II. This is the single most tested calculation.

Pattern 3: Coefficient of a Specific Power

Set the \(x\)-power in \(T_{r+1}\) equal to the required value and solve \(r\). Verify that \(r\) is a non-negative integer — if not, that power does not exist in the expansion (this is itself a question pattern from 2014-I Q25). Appeared in 2010-II, 2011-I, 2015-II, 2018-II.

Pattern 4: Identity / Equality of Coefficients

Questions like \(\binom{20}{n+2} = \binom{20}{n-2}\) or \(\binom{28}{2r} = \binom{28}{2r-4}\). Use \(\binom{n}{a} = \binom{n}{b} \implies a = b\) or \(a + b = n\). Always check both cases. Appeared in 2013-I, 2018-I, 2019-I, 2022-I.

Pattern 5: Number of Non-Zero Terms

In simplified expansions like \((x+a)^{n} + (x-a)^{n}\), odd-power terms cancel and only even-power terms survive. Non-zero count \(= \lfloor n/2 \rfloor + 1\). For \((1 + 2\sqrt{3}\, x)^{n} + (1 - 2\sqrt{3}\, x)^{n}\) with \(n = 11\) (odd): non-zero terms \(= (11+1)/2 = 6\). Appeared in 2017-I and 2018-I.

Pattern 6: Divisibility via Binomial Expansion

Express the base as \(1 + k\) or \(a + 1\), expand, and show that terms beyond the first two are divisible by \(k^{2}\). Remainders when divided by \(k^{2}\) come from the first two terms only. This pattern covers 2020-I (divisibility by 2000), 2024-II (\(7^{n} - 6n \bmod 36\)), and 2025-I and 2025-II (modular arithmetic with \(5^{99} \bmod 13\) and \((8 + 3\sqrt{7})^{100}\)).

Preparation Strategy

Week 1 — Formula Fluency

Write the general term \(T_{r+1}\) 20 times from memory with different \((a, b, n)\) combinations. Practice setting up the \(x\)-power equation and solving for \(r\). Do 10 term-independent-of-\(x\) problems in a single sitting to build speed. You should be able to write the general term of any binomial expression in under 15 seconds.

Week 2 — Coefficient Sums and Properties

Memorise the six key results: sum \(= 2^{n}\), even sum \(=\) odd sum \(= 2^{n-1}\), Pascal's identity, \(\binom{n}{r} = \binom{n}{n-r}\), middle-term rule, and the number-of-terms formula for \((a+b+c)^{n}\). Then solve all the coefficient-sum PYQs from 2010–2023 in a timed session (aim for 2 minutes per question).

Week 3 — Direction Sets and Log Problems

Practice the direction sets from 2014-I (Qs 15–17, Qs 22–26) and 2022-II (Qs 69–71) under exam conditions — read all sub-questions before solving the first. For log-inside-binomial problems, treat \(\log x\) as a constant factor and isolate the \(x\)-power separately. The 2023-I log problem (\(x = 10\)) is the template.

Week 4 — Divisibility and Mathematical Induction

For divisibility questions, the standard move is: express the base as \((1 + k)\), expand, and argue that everything beyond the first two terms has a factor of \(k^{2}\). Practice with 2024-II Q83 (\(7^{n} - 6n \bmod 36\)) and 2025-II Q89 (\(5^{99} \bmod 13\) using \(5^{4} \equiv 1 \pmod{13}\)). These take 3–4 minutes but are worth full marks if you know the method.

A realistic target for NDA Mathematics: score 4 out of 5 questions on Binomial Theorem. Given the predictability of question patterns from 2010 to 2025, this is achievable with 3–4 weeks of focused practice. Pair this topic with Sequences and Series and Permutations and Combinations — all three topics share the ⁿCᵣ framework and together account for roughly 10–12 marks per paper.

Use the mock tests at /mocks to simulate the direction-set format under time pressure. Full-length NDA maths mock tests help you practise the decision: when to attempt a direction set first (high marks available) and when to skip and return.

For background on the broader algebra section, visit the NDA Maths subject index. Topics like Quadratic Equations and Inequalities and Sets, Relations and Functions often appear alongside Binomial Theorem on the same paper and share preparation time efficiently.

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