The Binomial Theorem gives you a formula to expand (a + b)ⁿ without multiplying out every bracket by hand. For any positive integer n:
where \binom{n}{r} = \frac{n!}{r!(n-r)!} is the binomial coefficient — also called C(n, r) or ⁿCᵣ in Indian textbooks.
Think of it this way: you are choosing which of the n brackets contribute a b (and the rest contribute a). When r brackets give a b, you get aⁿ⁻ʳ bʳ. The number of ways to choose those r brackets from n is C(n, r). Add up all values of r from 0 to n and you have the full expansion.
Analogy. Imagine you have 10 identical coins. Each coin shows either Heads (contributes a) or Tails (contributes b). The term with exactly r tails is C(10, r) · a^{10-r} · b^r. The total expansion (a + b)^{10} just sums all possibilities from zero tails to ten tails. Pascal's triangle is then just a visual bookkeeping device — each entry is the sum of the two entries above it, matching the identity C(n, r) = C(n-1, r-1) + C(n-1, r).
Why NDA tests this. The NDA paper has a pattern of bundling 3–4 questions around a single setup — typically (√2 + 1)^n and its conjugate (√2 - 1)^n. You need to recognise the conjugate-pair trick, extract the integer part, and apply the multiplicative inverse idea. These are mechanical once you see the structure, and you will see this exact structure in recent PYQs below.
For a positive integer n, the expansion of (a + b)^n has n + 1 terms:
Key properties worth committing to memory:
n + 1C(n,0) + C(n,1) + \cdots + C(n,n) = 2^n (set a = b = 1)2^{n-1} (set a = 1, b = -1)C(n, r) = C(n, n-r)The (r+1)-th term of (a + b)^n is:
This is the workhorse formula. For "find the term containing x^k" questions, write T_{r+1}, expand the powers, collect the exponent of x, set it equal to k, solve for r. If r is not a non-negative integer, that term doesn't exist.
Middle term. When n is even, the middle term is T_{n/2 + 1}. When n is odd, there are two middle terms: T_{(n+1)/2} and T_{(n+3)/2}.
This is the technique NDA tested directly. Look at (\sqrt{2} + 1)^{10} and its conjugate (\sqrt{2} - 1)^{10}.
When you expand both using the binomial theorem:
Now add them. For odd values of r, the (-1)^r = -1 cancels the corresponding term from the first expansion. For even values of r, the terms double:
Since r is even, 10 - r is also even, so (\sqrt{2})^{10-r} = 2^{(10-r)/2} — a rational number. Every term in this sum is a rational (in fact, integer) number. The full sum is therefore an integer.
Extracting the integer part. Define u + f = (\sqrt{2}+1)^{10} where u is the integer part and 0 < f < 1. Define v = (\sqrt{2}-1)^{10}.
Since \sqrt{2} - 1 \approx 0.414, we have 0 < \sqrt{2}-1 < 1, so 0 < (\sqrt{2}-1)^{10} < 1, meaning 0 < v < 1.
We showed u + f + v = N (an integer). Since u is already an integer, f + v must be an integer. Since 0 < f < 1 and 0 < v < 1, we have 0 < f + v < 2. The only integer in this open interval is 1. Therefore f + v = 1.
Here is a neat identity: (\sqrt{2}+1)(\sqrt{2}-1) = 2 - 1 = 1. These two numbers are multiplicative inverses of each other.
Therefore:
Since v = (\sqrt{2}-1)^{10}, we get (\sqrt{2}-1)^{20} = v^2. So the multiplicative inverse of (\sqrt{2}+1)^{20} is v^2.
To find u, compute (\sqrt{2}+1)^{10} + (\sqrt{2}-1)^{10}:
Using even-r terms (r = 0, 2, 4, 6, 8, 10):
| r | C(10,r) | (√2)^{10-r} | Term (×2 from addition) | |---|---------|-------------|--------------------------| | 0 | 1 | 2^5 = 32 | 2 × 1 × 32 = 64 | | 2 | 45 | 2^4 = 16 | 2 × 45 × 16 = 1440 | | 4 | 210 | 2^3 = 8 | 2 × 210 × 8 = 3360 | | 6 | 210 | 2^2 = 4 | 2 × 210 × 4 = 1680 | | 8 | 45 | 2^1 = 2 | 2 × 45 × 2 = 180 | | 10| 1 | 2^0 = 1 | 2 × 1 × 1 = 2 |
Sum: 64 + 1440 + 3360 + 1680 + 180 + 2 = 6726.
So u + f + v = 6726. Since f + v = 1, we get u = 6725.
When you add (A + B)^n + (A - B)^n and n is even, all irrational terms cancel because they pair up with opposite signs. The result is always rational (and often an integer when A and B are surd-integer combinations). Recognition test: if you see (√k ± 1)^n or (√k ± m)^n in a problem, immediately write the conjugate sum. Standard approach: manually verify irrationality term by term (~90 seconds). Pattern recognition: write conjugate sum directly and classify (~15 seconds).
For (√2 + 1) and (√2 - 1): their product is 1, so they are reciprocals. For any power n: (√2 - 1)^n = 1/(√2 + 1)^n. Whenever a question asks "multiplicative inverse of (√2 + 1)^k", write (√2 - 1)^k directly. Then rewrite in terms of v if v = (√2 - 1)^{10}: (√2 - 1)^{20} = v^2. This takes 8 seconds. Computing via rationalisation: ~40 seconds.
When 0 < conjugate base < 1, its power also lies in (0, 1). So both f (fractional part of the big expression) and v (the small conjugate power) lie in (0, 1). Their sum lies in (0, 2). If their sum must also be an integer, it can only be 1. Commit this squeeze: "two fractional parts that sum to an integer must sum to 1." Saves you from computing f and v separately. Step count: 3 steps vs 7+ steps by direct expansion.
For "find the term independent of x" in (x^a + x^{-b})^n: write T_{r+1} = C(n,r) · x^{a(n-r)} · x^{-br}. Set the exponent a(n-r) - br = 0, solve for r. Check r is a non-negative integer. If it is, compute C(n,r). Standard approach (expand partially to find the term): ~60 seconds. Substitution into general term and solving: ~20 seconds.
The sum of all coefficients in (1 + x)^n is 2^n. To verify your expansion is correct during a solve, sum the computed coefficients — if they add to 2^n, the expansion is consistent. For the term-count sanity check: number of terms = n + 1. For n = 10, expect 11 terms. This catches errors in under 5 seconds, where re-expanding from scratch takes 3+ minutes.
When you see a binomial theorem question in the exam hall, run this decision tree:
Step 1 — Identify the structure.
Is the base a conjugate pair like (√k ± m)? If yes, go to Step 2. If it is a straightforward (a + b)^n, go to Step 3.
Step 2 — Conjugate pair protocol.
(√k + m)^n + (√k - m)^n. Note that all irrational terms cancel.(√k - m) lies between 0 and 1. If yes, its power also lies in (0, 1).f + v = 1.Step 3 — General term protocol.
T_{r+1} = C(n, r) · a^{n-r} · b^r.x, coefficient of x^k).r.C(n, r) and the numerical coefficient.Step 4 — Sanity check. For coefficient questions, verify r is a non-negative integer not exceeding n. For integer-part questions, verify your conjugate sum is indeed an integer by checking that all irrational pieces cancel.
Why this question: This is the anchor question of a 4-part linked set. Understanding whether u + v + f and f + v are integers tests whether you know the conjugate cancellation technique.
Solving path: Expand (\sqrt{2}+1)^{10} + (\sqrt{2}-1)^{10}. All odd-r terms cancel (irrational parts). Even-r terms all give integer contributions (since (\sqrt{2})^{even} = 2^{integer}). The full sum is an integer N, so u + f + v = N — Statement I is correct. Since u is an integer, f + v = N - u is also an integer — Statement II is correct. Answer: Both I and II.
Why this question: Tests whether you can connect the conjugate product identity (\sqrt{2}+1)(\sqrt{2}-1) = 1 to a question phrased as "multiplicative inverse."
Solving path: Since (\sqrt{2}+1)(\sqrt{2}-1) = 1, raising both sides to the power 20 gives (\sqrt{2}+1)^{20} \cdot (\sqrt{2}-1)^{20} = 1. So the multiplicative inverse of (\sqrt{2}+1)^{20} is (\sqrt{2}-1)^{20}. Now v = (\sqrt{2}-1)^{10}, so (\sqrt{2}-1)^{20} = v^2. Answer: v^2.
Why this question: Tests the fractional-part squeeze. Many students try to compute f and v separately, which is slow and error-prone.
Solving path: We know f + v is an integer (from the previous part). Also \sqrt{2} - 1 \approx 0.414, so 0 < (\sqrt{2}-1)^{10} < 1, meaning 0 < v < 1. Combined with 0 < f < 1, we get 0 < f + v < 2. The only integer in (0, 2) is 1. Therefore f + v = 1. Answer: 1.
Why this question: This is the payoff question. It requires computing the conjugate sum explicitly and then using f + v = 1 to extract u. The computation is the most involved step in the set.
Solving path: Compute (\sqrt{2}+1)^{10} + (\sqrt{2}-1)^{10} by summing only the even-r terms (since odd terms cancel):
r = 0: C(10,0) \cdot 2^5 = 32, doubled: 64r = 2: C(10,2) \cdot 2^4 = 45 \cdot 16 = 720, doubled: 1440r = 4: C(10,4) \cdot 2^3 = 210 \cdot 8 = 1680, doubled: 3360r = 6: C(10,6) \cdot 2^2 = 210 \cdot 4 = 840, doubled: 1680r = 8: C(10,8) \cdot 2^1 = 45 \cdot 2 = 90, doubled: 180r = 10: C(10,10) \cdot 1 = 1, doubled: 2Total: 64 + 1440 + 3360 + 1680 + 180 + 2 = 6726.
So u + f + v = 6726. Since f + v = 1, we get u = 6725. Answer: 6725.
Wrong index on general term. The (r+1)-th term is T_{r+1} = C(n,r) \cdot a^{n-r} \cdot b^r. Students routinely write a^r \cdot b^{n-r} by swapping. Always fix a as the first term and b as the second.
Forgetting that the conjugate base must be checked for magnitude. The squeeze argument f + v = 1 only works because 0 < \sqrt{2}-1 < 1. If the conjugate base were greater than 1 (e.g., \sqrt{3} - 1 \approx 0.73 is fine, but \sqrt{5} - 1 \approx 1.24 is not), then v > 1 and the fractional-part argument breaks. Always verify the conjugate is between 0 and 1 before applying the squeeze.
Computing both expansions separately instead of using their sum. For the conjugate-pair integer questions, the only efficient path is to add the two expansions. Students who expand each separately and then add waste 3–4 minutes and risk arithmetic errors.
Off-by-one in term counting. The n-th term is T_n = C(n, n-1) a b^{n-1}, not C(n,n) b^n. When asked for "the 4th term," write T_4, which has r = 3 in the formula. The off-by-one error is the most common source of wrong answers on general-term questions.
Using C(n, r) values from the wrong row of Pascal's triangle. For n = 10, use row 10 (starting 1, 10, 45, 120, 210, 252, ...). C(10, 4) = 210, not 120. Verify: C(10,3) = 120, C(10,4) = 210. Mixing these up changes the final answer significantly.
Treating u as the value of the full expression, not the integer part. u = 6725, not 6726. The full expression (\sqrt{2}+1)^{10} equals 6725 + f where f is the fractional part. The conjugate sum 6726 equals u + 1 because f + v = 1. Students who skip the fractional-part analysis pick 6726 directly — wrong by exactly 1.