General Term in Binomial Expansion

General Term in Binomial Expansion

The binomial expansion is (x + a) ⁿ = ⁿC₀ xⁿ + ⁿC₁ xⁿ⁻¹ a + ⁿC₂ xⁿ⁻² a² + … + ⁿCr xⁿ⁻r ar + … + ⁿCnx⁰aⁿ.

  • (r + 1)th term is ⁿCr xⁿ⁻r ar.
    ∴ Tr₊₁ = ⁿCr xⁿ⁻r ar.
  • General term for (x – a )ⁿ is Tr₊₁ = (-1)r. ⁿCr xⁿ⁻r.
  • General term for (1 + x)ⁿ is Tr₊₁ = ⁿCr xr.
  • In the binomial expansion of (x + a)ⁿ, r th term from end is (n – r + 2)th term from beginning.
  • If n is odd then no .of terms in {(x + a)ⁿ + (x – a)ⁿ}, {(x + a )ⁿ – (x – a)ⁿ}. Have equal no. of terms =\(\left( \frac{n+1}{2} \right)\).
  • If n is odd then {(x + a)ⁿ + (x – a)ⁿ has \(\left( \frac{n}{2}+1 \right)\) terms {(x + a)ⁿ – (x – a)ⁿ} has (n/2) terms.


Step I: Write Tr ₊₁ and Tr from the given expansion.

Step II: Find \(\frac{{{T}_{r+1}}}{{{T}_{r}}}\)

Step III: put \(\frac{{{T}_{r+1}}}{{{T}_{r}}}>1\)

Step IV: Solve the inequality in step III for r to get an inequality of the form r < m or r > m. If m is an integer, then two are the greatest terms. If m is in magnitude and these two are the greatest terms. If m is not an integer, then obtain the integral part of m. say k, in this case, (k + 1)th term is the greatest term

Middle term in a binomial expansion:

  • If n is even. Then middle term of (x+ a)ⁿ is \({{\left( \frac{n}{2}+1 \right)}^{th}}\) term.
  • If n is odd. Then Middle terms and \({{\left( \frac{n+1}{2} \right)}^{th}}\) and \({{\left( \frac{n+3}{2} \right)}^{th}}\) terms.

Largest binomial coefficient: The largest among ⁿC₀, ⁿC₁, … ⁿCn is (are)

a) If n is event integer then \(^{n}{{C}_{\left( \frac{n}{2} \right)}}\) is largest

b) If n is odd integer \(^{n}{{C}_{\left( \frac{n-1}{2} \right)}}\), \(^{n}{{C}_{\left( \frac{n+1}{2} \right)}}\) both are greatest numbers

Examples: Find the middle terms in the expansion \({{\left( 3-\frac{{{x}^{3}}}{6} \right)}^{7}}\).

Solution: If n is odd middle terms are \({{\left( \frac{n+1}{2} \right)}^{th}}\) and \({{\left( \frac{n+3}{2} \right)}^{th}}\).

Given that \({{\left( 3-\frac{{{x}^{3}}}{6} \right)}^{7}}\)

n = odd = 7

Middle terms =  \({{\left( \frac{7+1}{2} \right)}^{th}}\) and \({{\left( \frac{7+3}{2} \right)}^{th}}\)

Middle terms are 4th and 5th

Given that \({{\left( 3-\frac{{{x}^{3}}}{6} \right)}^{7}}\)

T₄ = T₃₊₁.

b = \(\left( \frac{-{{x}^{3}}}{6} \right)\); r = 3.

T₄ = ⁷C₃ (3)⁷⁻³ \({{\left( \frac{-{{x}^{3}}}{6} \right)}^{3}}\)

= \(\frac{7!}{3!4!}\times 81\times {{\left( -1 \right)}^{3}}{{\left( \frac{-{{x}^{3}}}{6} \right)}^{3}}\)

T₄ = \(\frac{-105}{8}{{x}^{9}}\) … (1)

T₅   = T₄₊₁

b = \(\left( \frac{-{{x}^{3}}}{6} \right)\); r = 4

T₅ = ⁷C₄ (3)⁷⁻⁴ \({{\left( \frac{-{{x}^{3}}}{6} \right)}^{4}}\)

= \(\frac{7!}{3!4!}\times {{\left( 3 \right)}^{3}}{{\left( -1 \right)}^{4}}{{\left( \frac{-{{x}^{3}}}{6} \right)}^{4}}\)

T₅ = \(\frac{-35}{48}{{x}^{12}}\) … (2)

The middle terms are T₄ = \(\frac{-105}{8}{{x}^{9}}\) and T₅ = \(\frac{-35}{48}{{x}^{12}}\).