# 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.

Algorithm:

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 = $$\frac{-35}{48}{{x}^{12}}$$.