# Binomial expansion

The binomial theorem is used to calculate the expansion (x + y)n without carrying out direct multiplication. In the expansion x and y are real numbers and n is an integer.

For all positive integers n, the binomial (x + y) can be expanded:

(x + y)n = xn + nC1 x(n-1) and + nC2 x(n-2) y2 + nC3 x(n-3) y3 +… + Andn

Where the coefficients nCr appearing on the binomial expansion they are called binomial coefficients; these can also be expressed as  . To calculate them, the combinatorial formula is used:

nCr = n! / [(n - r)! ∙ r! ]

## Basic binomial expansions

• (x + y)0 = x0 = 1
• (x + y)1= x + {2! / [(2 - 1)! ∙ 1! ]} · X1y = x + y
• (x + y)2 = x2 + {2! / [(2-1)! ∙ 1! ]} · X1and + {2! / [(2 - 2)! ∙ 2! ]} · X0 y2 = x2 + 2xy + y2
• If n is a rational number and x is a real number such that | x | <1 then
• (1 + x)n = 1 + nx + nC2 x2 + ⋯ nCr xr + ⋯ ∞
• (1 + x)-n = 1 - nx + nC2 x2 + nC3 x3 + ⋯ ∞
• (1 - x)n = 1 - nx + nC2 x2 + nC3 x3 + ⋯ ∞
• (1 - x)-n = 1 + nx + nC2 x2 + ⋯ nCr xr + ⋯ ∞

Examples:

1. Solve (x + 20)3.

Since n = 3, we have to:

(x + y)3 = x3 +3! / [(3 - 1)! ∙ 1!] · X2and + 3! / [(3 - 2)! ∙ 2!] · X1y2 + 3! / [(3 - 3)! ∙ 3!] · X0 y3

= x3 + 3x2y + 3xy2 + and3

Now, as y = 20, we get

(x + 20)= x3 + 3x2(20) + 3x (20)2 + (20)=

= x3 + 60x2 + 1200x + 8000

1. Determine the binomial coefficients of the expansion (x + 2)6 using the binomial theorem.

Comparing (x + 2)6 with (x + y)n, we determine that y = 2 and n = 6. Therefore:

(x + y)6 = x6 + 6! / [(6 - 1)! ∙ 1!] · X5and + 6! / [(6 - 2)! ∙ 2!] · X4y2 + 6! / [(6 - 3)! ∙ 3!] · X3y3 + 6! / [(6 - 4)! ∙ 4!] · X2y4 + 6! / [(6 - 5)! ∙ 5!] · X1y5 + 6! / [(6 - 6)! ∙ 6!] · X0y6 =

= x6 + 6x2y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + and6

Thus:

(x + 2)6 = x6 + 6x2(2) + 15x4(2)2 + 20x3(2)3 + 15x2(2)4 + 6x (2)5 + (2)=

= x6 + 12x2 + 60x4 + 160x3 + 240x2 + 192x + 64

In consecuense:

• X coefficient6 = 1
• X coefficient5 = 12
• X coefficient4 = 60
• X coefficient3 = 160
• X coefficient2 = 240
• Coefficient of x = 192
• Constant = 64