| Binomial Expression:|
Def: An expression consisting of two terms is called a binomial expression, e.g, x+a, 2x+3y, 5x2-6y2, 2x-1/3x are all binomial expressions.
Binomial theorem helps us to expand any power of a given binomial expression. In this chapter, we propose to find the expansion of (x+a)n where n € +I. Here n is called the power or the index of the binomial expression.
Development of binomial expansion
By actual multiplication, we may obtain the following expansion.
(x+y)1 = x+y
(x+y)2 = x2 + 2xy + y2 = 2C0x2 + 2C1xy + 2C2y 2
(x+y)3 = x3+ 3x2y + 3xy2 + y3 = 3C0.x3 + 3C1. x2y + 3C2 .xy2 + 3 C2.y3
(x+y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
= 4C0x4 + 4C1x3y + 4C2x2y2 + 4C3xy3 + 4Cy4
(x+y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
= 5C0x5 + 5C1x4y + 5C2x3y2 + 5C3x2y3 + 5C4xy4 + 5C4xy4 + 5Cxy4xy4 + 5C5y5
A careful observation of these expansions shows that (x+y)n where (n = 1, 2, 3, 4, 5) when expanded has the following properties.
The co-efficients form a certain pattern as shown below:
Inspection will show that each term in the table is derived by adding together the two terms in the line above, which lie on either side of it. Thus in the line n = 5, the term 10 is found by adding together the terms 4 and 6 in the line n = 4
The co-efficients in combinatorial form called binomial co-efficients may be rewritten as
Continuing the above process, it is easily seen that last but one term [i.e, nth term] of the exapansion = nCn-1 = nxyn-1
- First term = nC0xn = xn
- Second term = nC1xn-1y = nxn-1y
- Third term = nC2xn-2y2 = [n(n-1)]/(1x2)xn-2y2
- Fourth term = nC3xn-3y3 = [n(n-1)(n=2)]/(1x2x3)xn-3y3
- Fifth term = nC5xn-y4 = [n(n-1)(n-2)(n-3)]/(1x2x3x4)xn-4y4
Last term = [i.e., (n+1)th term] = nCnyn = yn
(x+y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + ............ + nCrxn-ryr + .............. + nCn-1xyn-1 + nCnyn..........1
= xn + nxn-1y + [n(n-1)]/(1x2)xn-2y2 + [n(n-1)(n-2)]/(1x2x3)xn-3y3 + ............... +yn.
Example: Expand (1+4x)5
Solution: (1+4x)5 = 1 + 5C14x + 5C2(x)2 + 5C3(4x)3 + 5C4(4x)4 + (4x)5
= 1 + 5C14x + 5C2(4x)2 + 5C2(4x)3
+ 5C1(4x) + (4x)5 [because 5C3 = 5C3; 5C4 = 5C1]
= 1+5x4x + (5x4)/(1x2)x16x2 + (5x)/(1x2) x6x2 + 5x256x4 + 1024x5
Directions: Expand the following.