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### High School Mathematics - 27.5 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 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 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 (999)3 using binomial theorem Answer: Q 2: What is the number of terms in the expansion of [(x-2y)2]2]?151020 Q 3: Find the value of (1.004)8 using the binomial theorem.Answer: Q 4: Expand (2+x+x2)3Answer: Q 5: Find the value of (0.998)8 using the binomial theorem.Answer: Q 6: Expand (x-y/2)4Answer: Q 7: Expand (2+x+x2)3Answer: Q 8: Expand (3+2x2)4Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!

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