EAPCET Mathematics · Intermediate second year · Paper IIA

Binomial Theorem

Binomial Theorem expands (a + b)ⁿ for positive integers and (1 + x)ⁿ for rational n. Questions centre on the general term, middle terms, coefficients and approximations.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • General term: Tr+1=nCr an−rbrT_{r+1} = {}^{n}C_{r}\, a^{n-r} b^{r}
  • C0+C1+⋯+Cn=2nC_{0} + C_{1} + \dots + C_{n} = 2^{n}
  • C0+C2+C4+⋯=C1+C3+C5+⋯=2n−1C_{0} + C_{2} + C_{4} + \dots = C_{1} + C_{3} + C_{5} + \dots = 2^{n-1}
  • Middle term: Tn/2+1T_{n/2 + 1} when nn is even; T(n+1)/2T_{(n+1)/2} and T(n+3)/2T_{(n+3)/2} when nn is odd
  • (1+x)−1=1−x+x2−x3+…(1 + x)^{-1} = 1 - x + x^{2} - x^{3} + \dots for ∣x∣<1|x| < 1
  • For small xx, (1+x)n≈1+nx(1 + x)^{n} \approx 1 + nx
  • ∑r⋅nCr=n⋅2n−1\sum r \cdot {}^{n}C_{r} = n \cdot 2^{n-1} and ∑r2⋅nCr=n(n+1)⋅2n−2\sum r^{2} \cdot {}^{n}C_{r} = n(n + 1) \cdot 2^{n-2}
  • ∑r=0nnCrr+1=2n+1−1n+1\displaystyle\sum_{r=0}^{n} \dfrac{{}^{n}C_{r}}{r + 1} = \dfrac{2^{n+1} - 1}{n + 1}
  • C02+C12+⋯+Cn2=2nCnC_{0}^{2} + C_{1}^{2} + \dots + C_{n}^{2} = {}^{2n}C_{n}
  • C0Cr+C1Cr+1+⋯+Cn−rCn=2nCn−rC_{0}C_{r} + C_{1}C_{r+1} + \dots + C_{n-r}C_{n} = {}^{2n}C_{n-r}; in particular C0C1+C1C2+⋯+Cn−1Cn=2nCn−1C_{0}C_{1} + C_{1}C_{2} + \dots + C_{n-1}C_{n} = {}^{2n}C_{n-1}
  • Vandermonde: nC0⋅mCr+nC1⋅mCr−1+⋯+nCr⋅mC0=m+nCr{}^{n}C_{0} \cdot {}^{m}C_{r} + {}^{n}C_{1} \cdot {}^{m}C_{r-1} + \dots + {}^{n}C_{r} \cdot {}^{m}C_{0} = {}^{m+n}C_{r}
  • General term of (1−x)−m(1 - x)^{-m}: m+r−1Cr xr{}^{m+r-1}C_{r}\, x^{r}; of (1+x)−m(1 + x)^{-m}: (−1)r m+r−1Cr xr(-1)^{r}\,{}^{m+r-1}C_{r}\, x^{r} (∣x∣<1)(|x| < 1)
  • Multinomial: the coefficient of x1a1x2a2⋯xkakx_{1}^{a_{1}} x_{2}^{a_{2}} \cdots x_{k}^{a_{k}} in (x1+x2+⋯+xk)n(x_{1} + x_{2} + \dots + x_{k})^{n} is n!a1! a2!⋯ak!\dfrac{n!}{a_{1}!\,a_{2}! \cdots a_{k}!}, with a1+⋯+ak=na_{1} + \dots + a_{k} = n
  • nCrnCr−1=n−r+1r\dfrac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \dfrac{n - r + 1}{r} and nCr=nr⋅n−1Cr−1{}^{n}C_{r} = \dfrac{n}{r} \cdot {}^{n-1}C_{r-1}

What this chapter covers

  • Binomial theorem for a positive integral index
  • General term and middle terms
  • Properties of binomial coefficients
  • Numerically greatest term
  • Binomial theorem for a rational index
  • Approximations

How it is usually asked

  • Find the coefficient of xᵏ in an expansion
  • Find the term independent of x
  • Evaluate sums of binomial coefficients
  • Approximate values such as √(1.02)

Understanding Binomial Theorem

For a positive integer n, (a + b)ⁿ expands into n + 1 terms, and the (r + 1)th term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. The coefficients ⁿC₀, ⁿC₁, …, ⁿCₙ read the same from both ends, rise to the middle and fall again, which is why the middle term or terms carry the largest binomial coefficient. When n is even there is one middle term; when n is odd there are two.

Coefficient questions all follow one routine: write the general term, collect the power of x it contains, set that power equal to the one required and solve for r. A term independent of x has power zero. Sums of coefficients come from substituting values: x = 1 gives C₀ + C₁ + … + Cₙ = 2ⁿ, and x = −1 shows the even-place and odd-place sums are equal, each 2ⁿ⁻¹.

For a negative or fractional index, (1 + x)ⁿ becomes an infinite series 1 + nx + n(n − 1)x²/2! + …, valid only when |x| < 1. Keeping the first two or three terms gives quick approximations such as √(1.02) ≈ 1.01. EAPCET questions ask for specific coefficients, independent terms, middle terms, coefficient sums and such approximations.

Worked example

EAPCET-style question

The term independent of xx in the expansion of (x2+2x)6\left(x^{2} + \dfrac{2}{x}\right)^{6} is

  1. (1)60
  2. (2)160
  3. (3)240
  4. (4)480

Solution

  1. General term: Tr+1=6Cr (x2)6−r(2x)r=6Cr 2rx12−3rT_{r+1} = {}^{6}C_{r}\,(x^{2})^{6-r}\left(\dfrac{2}{x}\right)^{r} = {}^{6}C_{r}\, 2^{r} x^{12-3r}.
  2. For the term independent of xx, 12−3r=012 - 3r = 0, so r=4r = 4.
  3. T5=6C4×24=15×16=240T_{5} = {}^{6}C_{4} \times 2^{4} = 15 \times 16 = 240.

Answer: 240

Shortcuts and results to remember

  • The kkth term from the end of (a+b)n(a + b)^{n} is the (n−k+2)(n - k + 2)th term from the beginning.
  • Remainder of aba^{b} on division by cc: write a power of aa as (multiple of c±1c \pm 1) and expand; every term except the last is a multiple of cc.
  • C02−C12+C22−⋯+(−1)nCn2=0C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - \dots + (-1)^{n}C_{n}^{2} = 0 if nn is odd and (−1)n/2 nCn/2(-1)^{n/2}\,{}^{n}C_{n/2} if nn is even.
  • For a negative or fractional index, (a+b)n=an(1+ba)n(a + b)^{n} = a^{n}\left(1 + \dfrac{b}{a}\right)^{n} is valid only when ∣ba∣<1\left|\dfrac{b}{a}\right| < 1.

More practice questions

Practice question 1

The remainder when 5995^{99} is divided by 13 is

  1. (1)8
  2. (2)5
  3. (3)1
  4. (4)12

Solution

  1. 599=5⋅(52)49=5⋅(26−1)495^{99} = 5 \cdot (5^{2})^{49} = 5 \cdot (26 - 1)^{49}
  2. Expanding, every term of (26−1)49(26 - 1)^{49} except (−1)49(-1)^{49} is a multiple of 13
  3. So 599≡5⋅(−1)=−5(mod13)5^{99} \equiv 5 \cdot (-1) = -5 \pmod{13}
  4. −5+13=8-5 + 13 = 8

Answer: 8

Practice question 2

The coefficient of a2b3c4a^{2}b^{3}c^{4} in the expansion of (a+b+c)9(a + b + c)^{9} is

  1. (1)1260
  2. (2)630
  3. (3)2520
  4. (4)84

Solution

  1. Multinomial coefficient =9!2! 3! 4!= \dfrac{9!}{2!\,3!\,4!}
  2. =3628802×6×24=362880288= \dfrac{362880}{2 \times 6 \times 24} = \dfrac{362880}{288}
  3. =1260= 1260

Answer: 1260

Common mistakes to avoid

  • Treating Tᵣ as the general term ⁿCᵣ aⁿ⁻ʳ bʳ; that expression is Tᵣ₊₁, the (r + 1)th term.
  • Forgetting to raise the constant in the second term to the power r, such as dropping the 2ʳ or the (−1)ʳ in the coefficient.
  • Counting the middle term as the (n/2)th; for even n there are n + 1 terms, so the middle one is the (n/2 + 1)th.
  • Using the rational-index expansion of (1 + x)ⁿ when |x| ≥ 1, where the series does not converge.

How to prepare it

Write the general term once, collect the power of x, and solve for r — most coefficient questions are just that.

For coefficient-sum questions, substitute x = 1 or x = −1 in the expansion.

Once the formulas feel familiar, test them the way the exam does: practise Binomial Theorem questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Binomial Theorem chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Binomial Theorem part of the EAPCET Mathematics syllabus?
Yes. Binomial Theorem is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Binomial Theorem for EAPCET?
Write the general term once, collect the power of x, and solve for r — most coefficient questions are just that. For coefficient-sum questions, substitute x = 1 or x = −1 in the expansion.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Binomial Theorem?
Treating Tᵣ as the general term ⁿCᵣ aⁿ⁻ʳ bʳ; that expression is Tᵣ₊₁, the (r + 1)th term. Forgetting to raise the constant in the second term to the power r, such as dropping the 2ʳ or the (−1)ʳ in the coefficient. Counting the middle term as the (n/2)th; for even n there are n + 1 terms, so the middle one is the (n/2 + 1)th. Using the rational-index expansion of (1 + x)ⁿ when |x| ≥ 1, where the series does not converge.