# 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 rth term
= an–r+1 xr–1 [{n(n–1) (n – 2) ... (n – r + 2)} ÷ (r – 1)!].
Here r = 3 and n = 6.
So 3rd term of (3 + y)6 = 36 – 3 + 1 . y3 – 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
= an – r + 1 xr – 1 [{n (n – 1) (n – 2) ... (n – r + 2)} ÷ (r – 1)!].
z4 will come in 5th term.
Hence we have to find the 5th term of the expansion.
Here r = 5 and n = 8.
So 5th term of (5 + z)8 =5 8 – 5 + 1. z5 – 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 rth 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 2nd term of (p + 2)6 = p6 – 2 + 1 . 26 – 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