Multiplying (a + b)n by hand for large n is painful — the Binomial Theorem gives every term in one shot. Pair it with logarithms, which turn multiplication into addition and powers into products, and you have two of the most reliable scoring areas in NDA Maths. This lesson builds both from scratch with exam-ready formulas and shortcuts.
Why Binomial and Logarithms Matter for NDA
Almost every NDA paper carries questions from the Binomial Theorem and from logarithms. They are formula-driven, which is great news: once you know the standard results, you can answer in under a minute without any heavy calculation. There is no lengthy proof to reproduce and no diagram to draw — you simply pick the right formula and substitute.
Binomial questions usually ask for a particular term, a coefficient, the middle term, or the term free of x. Logarithm questions test the basic laws, base change, and simple exponential equations. Both reward memory plus a little practice. Because the answers are exact numbers, you can also verify your work quickly and avoid silly errors under time pressure.
Another reason these chapters are loved by toppers is that the questions are rarely twisted. Once you recognise the type — “find the coefficient”, “find the middle term”, “simplify the log” — the method is fixed. So your job is to build a small bank of formulas and apply them mechanically.
These topics are direct-formula chapters. Marks here are the cheapest in the whole Maths paper — do not skip them.
The Binomial Theorem Statement
For any positive integer n, the expansion is:
(a + b)n = nC0 an + nC1 an−1b + nC2 an−2b2 + … + nCn bn
where nCr = n! ÷ [ r! (n − r)! ] are the binomial coefficients.
Key features of the expansion
- There are exactly (n + 1) terms.
- The power of a decreases from n to 0; the power of b increases from 0 to n.
- In every term, the sum of the powers of a and b equals n.
- Coefficients are symmetric: nCr = nCn−r.
For (a − b)n, the signs alternate + − + −, because b is replaced by (−b). So in (a − b)n every odd-position term (the 2nd, 4th, 6th …) carries a minus sign while the even-position terms stay positive.
It also helps to remember Pascal's triangle, where each row gives the coefficients of (a + b)n and each number is the sum of the two numbers directly above it. For small n — say n = 2, 3 or 4 — reading the coefficients off Pascal's triangle is often faster than computing nCr by hand.
The General Term (Tr+1)
The single most useful binomial formula is the general term. It lets you write any term directly without expanding the whole thing.
Tr+1 = nCr an−r br
This is the (r + 1)th term of (a + b)n. Put r = 0 for the first term, r = 1 for the second, and so on.
To find the 5th term, use r = 4. To find the coefficient of x7, write the general term, simplify the power of x, set that power equal to 7, and solve for r.
The term number is always one more than r. If a question says “6th term”, immediately set r = 5.
Middle Term and Greatest Coefficient
The middle term depends on whether n is even or odd, because there are (n + 1) terms in total.
- If n is even: one middle term, the (n/2 + 1)th term.
- If n is odd: two middle terms, the ((n+1)/2)th and ((n+3)/2)th terms.
The greatest binomial coefficient sits at the middle. For even n it is nCn/2; for odd n the two equal greatest coefficients are nC(n−1)/2 and nC(n+1)/2.
Do not confuse the greatest coefficient (a property of the numbers nCr) with the greatest term (which also depends on the values of a and b). They are different questions.
Term Independent of x
A favourite NDA question asks for the term independent of x (the constant term) in an expansion like (x2 + 1/x)n.
Method
- Write the general term Tr+1.
- Collect all powers of x into a single exponent.
- Set that exponent equal to 0 (since x0 = 1).
- Solve for r, then substitute back to get the constant.
If solving gives a non-integer r, then no term independent of x exists — the answer is 0 or “does not exist”.
The same four-step method works for any “coefficient of xk” question: instead of setting the exponent to 0, set it equal to k. This is why mastering the general term unlocks almost the entire binomial portion of the syllabus.
Useful Binomial Identities
Putting special values of a and b into (a + b)n produces handy results that appear directly in objective papers.
- Sum of all coefficients: put a = b = 1 → nC0 + nC1 + … + nCn = 2n.
- Put a = 1, b = −1 → nC0 − nC1 + nC2 − … = 0.
- Hence sum of even-position coefficients = sum of odd-position coefficients = 2n−1.
Also note (1 + x)n = nC0 + nC1x + nC2x2 + … + nCnxn, which is the most common form used in NDA problems. From it, the coefficient of xr is simply nCr, so a question like “coefficient of x3 in (1 + x)8” is answered in one step as 8C3 = 56.
When you see (1 + x)n, do not expand. Just read off the coefficient you need as nCr — this saves precious seconds in the exam hall.
Worked Example: Binomial
Find the term independent of x in the expansion of (x2 + 1/x)6.
So the term independent of x is 15.
Notice how the general term did all the work — no full expansion was needed.
Logarithms: Core Idea and Laws
A logarithm simply answers the question: “To what power must the base be raised to get this number?” In other words, a log is just an exponent written in a different way. This single insight makes the whole chapter click into place.
If ax = N (with a > 0, a ≠ 1, N > 0), then logaN = x.
Example: 23 = 8, so log28 = 3.
Two standard bases
- Common logarithm: base 10, written log N. Used in most numerical work and log tables.
- Natural logarithm: base e (≈ 2.718), written ln N. Common in calculus and growth problems.
The number inside a log (the argument) must be strictly positive. log of 0 or of a negative number is undefined. Also the base can never be 1, because 1 raised to any power is always 1.
Laws of logarithms
These laws convert hard operations into easy ones — multiplication becomes addition and powers become products — and they are tested almost every year.
- Product: loga(MN) = logaM + logaN
- Quotient: loga(M/N) = logaM − logaN
- Power: loga(Mp) = p × logaM
- loga1 = 0 and logaa = 1
Change of base
logaN = logbN ÷ logba
A useful corollary: logab × logba = 1, i.e. logab = 1 ÷ logba.
Use the power law to pull exponents down. Exponential equations like 2x = 7 are solved instantly by taking log of both sides: x = log 7 ÷ log 2.
Worked Example: Logarithms
If log102 = 0.3010, find log105.
So log105 = 0.6990. The same trick gives log 4 = 2 log 2 = 0.6020.
Knowing just log 2 = 0.3010 and log 3 = 0.4771 lets you compute the logs of 4, 5, 6, 8, 9, 10 and more — memorise these two values.
For instance, log 6 = log(2 × 3) = 0.3010 + 0.4771 = 0.7781, and log 9 = 2 log 3 = 0.9542. By combining the product, quotient and power laws with these two numbers, you can build a surprisingly large table of logarithms in your head — a skill the NDA paper rewards directly.
Common Mistakes to Avoid
Most lost marks in these two chapters come from a handful of avoidable errors. Read this list twice before your exam — spotting your own habit here can save several marks.
Binomial slips
- Forgetting that the term number is r + 1, not r, so the 6th term needs r = 5.
- Dropping the negative sign in (a − b)n for odd-position terms.
- Mixing up greatest coefficient with greatest term; the first depends only on n, the second also on a and b.
- Writing n! ÷ r! and forgetting the (n − r)! in the denominator of nCr.
Logarithm slips
- Writing log(M + N) = log M + log N — false. The product law uses M × N, not M + N.
- Treating (log M)2 the same as log M2. They are different: log M2 = 2 log M, but (log M)2 is the whole log squared.
- Forgetting the domain restriction (argument > 0) and ticking an answer that is actually undefined.
logaM ÷ logaN is not loga(M/N). The quotient law gives a subtraction (log M − log N), while a ratio of two logs is the change-of-base form.
Previous-Year Style Question
Q. What is the coefficient of x5 in the expansion of (1 + x)10?
Answer: The general term is Tr+1 = 10Cr xr. For x5, set r = 5, giving coefficient 10C5 = 252.
Q. The value of log28 + log39 is?
Answer: log28 = log223 = 3 and log39 = log332 = 2, so the sum is 3 + 2 = 5.
Quick Revision
- (a + b)n has (n + 1) terms; general term Tr+1 = nCr an−r br.
- Even n → one middle term; odd n → two middle terms.
- Sum of all binomial coefficients = 2n.
- For the constant term, set the power of x to 0 and solve for r.
- Log laws: product → add, quotient → subtract, power → multiply.
- Change of base: logaN = logbN ÷ logba; and logab = 1 ÷ logba.
Drill five binomial and five log questions daily for a week and these become guaranteed marks in your NDA Maths paper.
Frequently asked questions
How many terms are there in the expansion of (a + b)^n?
There are exactly (n + 1) terms. For example, (a + b)^5 has six terms. The powers of a fall from n to 0 while the powers of b rise from 0 to n.
What is the general term in the binomial theorem?
The general term is T(r+1) = nCr a^(n-r) b^r. It gives any term directly without expanding the whole binomial, which is why it is the most used formula in this chapter.
How do I find the middle term of a binomial expansion?
If n is even there is one middle term, the (n/2 + 1)th term. If n is odd there are two middle terms, the ((n+1)/2)th and ((n+3)/2)th terms.
Why can't a logarithm have a negative or zero argument?
A logarithm asks what power of a positive base gives the number. Since a positive base raised to any real power is always positive, the argument must be strictly greater than zero.
What is the change of base formula and when is it used?
log_a(N) = log_b(N) / log_b(a). It is used to convert a logarithm to a base your calculator or table supports, usually base 10 or base e, and to simplify mixed-base expressions.
Are Binomial Theorem and Logarithms important for NDA?
Yes. Both are formula-driven topics that appear almost every year and can be solved quickly, making them among the highest scoring areas in the NDA Maths paper for well-prepared students.
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