Q.Using binomial theorem, evaluate (96)3.
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The Problem That Started It All
Imagine you have to expand (x+y)2. That's easy: x2+2xy+y2. Now try (x+y)3: x3+3x2y+3xy2+y3. Still manageable.
But what about (x+y)10? Or (x+y)100? Multiplying it out term by term would take forever. There has to be a pattern — and there is.
The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of (x+y)n looks like, without ever having to multiply.
The Pattern You Already Know
Look at the expansions you already know:
| Power | Expansion |
|---|---|
| (x+y)0 | 1 |
| (x+y)1 | x+y |
| (x+y)2 | x2+2xy+y2 |
| (x+y)3 | x3+3x2y+3xy2+y3 |
| (x+y)4 | x4+4x3y+6x2y2+4xy3+y4 |
Notice three things:
- The powers of x decrease from n down to 0, while the powers of y increase from 0 up to n. In every term, the exponents add to n.
- The coefficients — 1, 4, 6, 4, 1 for n=4 — follow a famous pattern called Pascal's triangle.
- The number of terms is always n+1.
Pascal's triangle: each number is the sum of the two numbers directly above it.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
The Precise Statement
Binomial Theorem: For any positive integer n,
(x+y)n=∑k=0n(kn)xn−kyk
where (kn)=k!(n−k)!n! is called the binomial coefficient.
Let's break that down.
The symbol ∑k=0n means "add up terms for k=0,1,2,…,n". For each k, the term is:
- (kn) — the coefficient (read as "n choose k")
- xn−k — x raised to the power n−k
- yk — y raised to the power k
So for n=4, the terms are:
| k | (k4) | x4−k | yk | Term |
|---|---|---|---|---|
| 0 | (04)=1 | x4 | y0 | 1⋅x4 |
| 1 | (14)=4 | x3 | y1 | 4x3y |
| 2 | (24)=6 | x2 | y2 | 6x2y2 |
| 3 | (34)=4 | x1 | y3 | 4xy3 |
| 4 | (44)=1 | x0 | y4 | 1⋅y4 |
Add them up: x4+4x3y+6x2y2+4xy3+y4. Exactly what we had.
Where Do Those Coefficients Come From?
The binomial coefficient (kn) counts how many ways you can choose k items from a set of n items. In the expansion, it counts how many ways you can pick k copies of y (and therefore n−k copies of x) when multiplying (x+y) by itself n times.
To compute (kn) quickly: start at n and multiply k decreasing numbers, then divide by k!.
Example: (37)=3⋅2⋅17⋅6⋅5=35.
The General Term
The k-th term (starting with k=0) in the expansion is:
General term: Tk+1=(kn)xn−kyk
This is the most useful part for exams. If someone asks "find the 5th term in (x+y)10", you set k=4 (because Tk+1 means k=4 gives the 5th term) and write:
T5=(410)x6y4
What If It's Not Just x and y? …
Concept: Binomial Theorem Expansion
We express 96 as a binomial sum that makes calculation easy: 96=100−4.
Now apply the binomial theorem to (100−4)3:
(100−4)3=(03)(100)3−(13)(100)2(4)+(23)(100)(4)2−(33)(4)3
Computing each term:
- (03)(100)3=1⋅1000000=1000000
- (13)(100)2(4)=3⋅10000⋅4=120000 …
Express 96 as 100−4 and expand (100−4)3 using the binomial theorem; the calculation becomes straightforward and gives 884736.
The binomial theorem is powerful when you need to compute powers of numbers that are close to convenient round numbers. Here, 96 is just 4 away from 100, so we can write 96=100−4 and expand (100−4)3 using the theorem.
The binomial theorem states that for any real numbers a and b, and positive integer n:
(a+b)n=∑k=0n(kn)an−kbk=(0n)an+(1n)an−1b+(2n)an−2b2+⋯+(nn)bn
For n=3, this becomes:
(a+b)3=a3+3a2b+3ab2+b3
Now let's apply this with a=100 and b=−4.
Step-by-step expansion:
-
Identify the terms. We have (100+(−4))3, so a=100, b=−4, and n=3.
-
Write out the binomial expansion:
(100−4)3=(03)(100)3+(13)(100)2(−4)+(23)(100)(−4)2+(33)(−4)3
-
Calculate each binomial coefficient:
- (03)=1
- (13)=3
- (23)=3
- (33)=1
-
Compute each term separately:
- First term: 1⋅(100)3=1000000
- Second term: 3⋅(100)2⋅(−4)=3⋅10000⋅(−4)=−120000 …
Showing the 12 most recent of 31 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.Let (2−x)9=a0+a1x+a2x2+……+a9x9. Then the value of a1+a2+a3+……+a8 is equal to (A) -511 (B) 510 (C) -512 (D) 512 (E) -510
›Reveal solutionSolution
Sum all coefficients (=1), then remove a0=512 and a9=−1 to get −510.
Substituting x=1 into (2−x)9=∑akxk:
a0+a1+⋯+a9=(2−1)9=1.
The endpoint coefficients are a0=29=512 (constant term) and a9=(−1)9=−1 (coefficient of x9). …
- KEAM 2026Set eng-2026-04174 marksMCQQ.If the coefficient of x3 in the binomial expansion of (2+x)n is 160, then the coefficient of x6 in the binomial expansion of (2−x2)n is (A) 160 (B) 320 (C) -160 (D) -320 (E) -960
›Reveal solutionSolution
The condition gives n=6; the x6 term of (2−x2)6 is (36)23(−1)3=−160.
The coefficient of x3 in (2+x)n is (3n)2n−3. Setting it to 160 and testing, n=6 works: (36)23=20⋅8=160. …
- KEAM 2026Set eng-2026-04184 marksMCQQ.If the coefficient of x4 in the binomial expansion of (4x+a)7 is −1120, then the value of a is equal to (A) 21 (B) 41 (C) 2−1 (D) 81 (E) 8−1
›Reveal solutionSolution
The x4 term gives 8960a3=−1120, so a=−21.
The general term of (4x+a)7 is 7Ck(4x)ka7−k. For x4 take k=4:
7C4(4)4a3=35⋅256⋅a3=8960a3.
Setting this equal to −1120: …
- KEAM 2026Set eng-2026-04194 marksMCQQ.The coefficient of x21 in the binomial expansion of (3x−3x1)4, is (A) 74 (B) 83 (C) 92 (D) 94 (E) 9−4
›Reveal solutionSolution
Write the general term, set the exponent of x equal to −2 to find k=3, then evaluate the coefficient: −94.
Concept. For (a+b)n the general (,(k+1)-th,) term is Tk+1=(kn)an−kbk. Here a=3x, b=−3x1, n=4.
Step 1 — general term.
Tk+1=(k4)(3x)4−k(−3x1)k=(k4)(−1)k34−k3−kx4−kx−k.
Combining powers,
Tk+1=(k4)(−1)k34−2kx4−2k. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.The coefficient of x8 in the expansion of (x2+1−x2)5+(x2−1−x2)5, is (A) −20 (B) −10 (C) −30 (D) −40 (E) 20
›Reveal solutionSolution
Adding (u+v)5+(u−v)5 with u=x2, v=1−x2 leaves only even powers of v: 2[u5+10u3v2+5uv4]. Substituting v2=1−x2 and collecting, the x8 term has coefficient −20.
Let u=x2 and v=1−x2, so v2=1−x2. Then
(u+v)5+(u−v)5=2[(05)u5+(25)u3v2+(45)uv4]=2[u5+10u3v2+5uv4].
Substitute v2=1−u (in u=x2):
u5=x10,
10u3(1−u)=10x6−10x8, …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If the 17th and 18th term in the expansion of (2+x)50 are equal, then the value of x is equal to (A) 1 (B) 2 (C) 4 (D) 6 (E) 8
›Reveal solutionSolution
Equating the 17th and 18th terms gives x=2⋅(1650)/(1750)=1.
In (2+x)50, Tk+1=(k50)250−kxk. The 17th term is k=16, the 18th is k=17:
(1650)234x16=(1750)233x17. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.If the coefficient of y3 in the binomial expansion of (2α−2y)8 is −7, then the value of α is equal to (A) 21 (B) 2 (C) 3 (D) 6 (E) 8
›Reveal solutionSolution
Pick the general term with y3, set its coefficient to −7, solve for α.
General term of (2α−2y)8: (r8)(2α)8−r(−2y)r. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.Let (1+ax)(1−2x)3=∑n=04anxn, where a is a constant. If a2=0, then the value of a is (A) 1 (B) 9 (C) 3 (D) 5 (E) 2
›Reveal solutionSolution
Expand (1−2x)3, collect the x2 term, set it to zero.
(1−2x)3=1−6x+12x2−8x3.
Multiplying by (1+ax), the x2 coefficient comes from 1⋅12x2 and ax⋅(−6x):
a2=12−6a. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.In the binomial expansion of (2x+α)8, the co-efficients of x2 and x3 are equal. Then the value of α is equal to (A) 2 (B) 41 (C) 4 (D) 21 (E) 3
›Reveal solutionSolution
Equating the x^2 and x^3 coefficients of (2x+alpha)^8 gives alpha = 4.
Concept and Intuition
In (2x + alpha)^8 the general term is C(8,k) (2x)^k alpha^(8-k), so the coefficient of x^k is C(8,k) 2^k alpha^(8-k). Setting the k=2 and k=3 coefficients equal isolates alpha.
Step-by-Step Solution
- Coefficient of x^2: C(8,2) 2^2 alpha^6 = 284alpha^6 = 112 alpha^6.
- Coefficient of x^3: C(8,3) 2^3 alpha^5 = 568alpha^5 = 448 alpha^5. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The constant term in the binomial expansion of (2x−x25)6 is (A) 4800 (B) 3200 (C) 5600 (D) 5400 (E) 6000
›Reveal solutionSolution
Set the exponent of x to zero and evaluate that term.
The general term of (2x−x25)6 is
(k6)(2x)6−k(−x25)k=(k6)26−k(−5)kx6−3k. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The coefficient of x10 in (1−x2)(1−x3)9 is (A) 9C4 (B) −9C6 (C) −9C4 (D) 9C6 (E) 0
›Reveal solutionSolution
(1−x3)9=∑(k9)(−1)kx3k; to make x10 need 3k=10 or 3k=8 — impossible — so coefficient =0.
Expand. (1−x3)9=∑k=09(k9)(−1)kx3k, giving exponents 0,3,6,9,… …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The constant term in (2x+3x21)10 is (A) 1285 (B) 1289 (C) 2565 (D) 2569 (E) 0
›Reveal solutionSolution
Set the exponent of x to zero: 210−r−2r=0⇒r=2; evaluate the term to get 2565.
General term. Tr+1=(r10)(2x)10−r(3x21)r=(r10)210−r13r1x210−r−2r. …
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