Binomial Expansion: Solved Examples

Example 1: Expand (5x – 4)¹⁰

Sol: (5x – 4)¹⁰ = ¹⁰C0 (5x)^10–0(–4)⁰ + ¹⁰C1 (5x)^10–1(–4)¹
+ ¹⁰C₂ (5x)^10–2(–4)² + ¹⁰C₃ (5x)^10–3(–4)³
+ ¹⁰C₄ (5x)^10–4(–4)⁴ + ¹⁰C₅ (5x)^10–5(–4)⁵
+ ¹⁰C₆ (5x)^10–6(–4)⁶ + ¹⁰C₇ (5x)^10–7(–4)⁷
+ ¹⁰C₈ (5x)^10–8(–4)⁸ + ¹⁰C₉ (5x)^10–9(–4)⁹
+ ¹⁰C₁₀ (5x)^10–10(–4)¹⁰

Example 2:Find the third term in the expansion of (3 + y)⁶

Sol: As expansion is of the form (a + x)ⁿ, so r^th term
= a^n–r+1 x^r–1 [{n(n–1) (n – 2) ... (n – r + 2)} ÷ (r – 1)!].
Here r = 3 and n = 6.
So 3^rd term of (3 + y)⁶ = 3^{6 – 3 + 1} . y^{3 – 1} . [(6x5)/2]
=3⁴ . y² . 15 = 1215 y²

Example 3:Find the co-efficient of z4 in the expansion of (5 + z)8.

Sol: As expansion is of the form (a + x)ⁿ, so rth term
= a^{n – r + 1} x^{r – 1} [{n (n – 1) (n – 2) ... (n – r + 2)} ÷ (r – 1)!].
z4 will come in 5^th term.
Hence we have to find the 5^th term of the expansion.
Here r = 5 and n = 8.
So 5^th term of (5 + z)⁸ =5 ^{8 – 5 + 1}. z^{5 – 1} .
= 5⁴ . z⁴ . 70 = 625x70x4 = 43750 z⁴
Hence coefficient is z⁴ is 43750.

Example 4: Find the co-efficient of p5 in the expansion of (p + 2)⁶.

Sol: As expansion is of the form (x + a)ⁿ, so r^th term
= x ^{n – r + 1} a ^{r – 1} [{n(n–1) (n – 2) ... (n – r + 2)} ÷ (r – 1)!].
So x⁵ will come when r = 2 and n = 6.
Hence we have to find the 2nd term of the expansion.
So r = 2 and n = 6.
So 2^nd term of (p + 2)⁶ = p^{6 – 2 + 1} . 2^{6 – 1} .
= p⁵ . 2⁵ . 6 = 192 p⁵
Hence coefficient of p⁵ is 192.

Example 5: For what value of x, the fifth term of the following expansion is equal to 105?

Sol: (-1)⁶ . ¹⁰c₄ (1/2√x)⁶(1/2)⁴ = 105 => x³ = ¹⁰c₄/105.2¹⁰
x = 1/8