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High School Mathematics - 2
7.5 Binomial Theorem

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
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:
Pascal's Triangle
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

  1. First term = nC0xn = xn
  2. Second term = nC1xn-1y = nxn-1y
  3. Third term = nC2xn-2y2 = [n(n-1)]/(1x2)xn-2y2
  4. Fourth term = nC3xn-3y3 = [n(n-1)(n=2)]/(1x2x3)xn-3y3
  5. Fifth term = nC5xn-y4 = [n(n-1)(n-2)(n-3)]/(1x2x3x4)xn-4y4
Continuing the above process, it is easily seen that last but one term [i.e, nth term] of the exapansion = nCn-1 = nxyn-1
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.
Q 1: Find the value of (10.1)5 using the binomial theorem.
Answer:

Q 2: Find the value of (999)3 using binomial theorem
Answer:

Q 3: Expand (3+2x2)4
Answer:

Q 4: Expand (2+x+x2)3
Answer:

Q 5: Find the value of (1.004)8 using the binomial theorem.
Answer:

Q 6: Expand (3x-y)4.
Answer:

Q 7: Find the value of (0.998)8 using the binomial theorem.
Answer:

Q 8: What is the number of terms in the expansion of [(x-2y)2]2]?
10
20
15

Question 9: This question is available to subscribers only!

Question 10: This question is available to subscribers only!


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