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Presentation transcript:

Mathematics

Binomial Theorem Session 2

Session Objectives

Session Objective Properties of Binomial Coefficients Binomial theorem for rational index — General term — Special cases 3. Application of binomial theorem — Divisibility — Computation and approximation

Properties of Binomial Coefficients The sum of the binomial coefficients in the expansion of (1+x)n is 2n i.e. where Cr = nCr Proof: Put x = 1 both sides we get For example 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 1 + 4 + 6 + 4 + 1 = 16 = 24

Properties of Binomial Coefficients 2. Sum of coefficients of the odd terms = Sum of the coefficients of the even terms in (1+x)n = 2n-1 i.e. Proof: Put x = –1 in above, we get Ask students to find 4c0+4c2+4c4 and 4c1+4c3 and show that it is 2^3 = 8 As sum of all the coefficients is 2n

Properties of Binomial Coefficients 3. Proof: For example:

Class Exercise - 1 Solution : If C0, C1, C2,... denote the binomial coefficients in the expansion of (1+x)n then prove that Solution :

Solution Cont.

Class Exercise - 2 Solution :

Solution Cont. = 0 + 0 + 0 = 0

Binomial theorem for Rational Index Let n be a rational number and x be real number such that |x| < 1, then Conditions of validity: if n is not a whole number Note that if n is a whole number then given expansion is same as that of binomial theorem for positive integral index i) |x| < 1 ii) Number of terms is infinite

General Term in (1+x)n, n  Q Expansion of (a + x)n for rational n Case1: Case2:

Class Exercise - 6 Write the first four terms in the expansion of For what values of x is this expansion is valid? Also, find the general term in this expansion. Solution :

Solution Cont. Validity if General term

Special Cases

Special Cases Note here that all these expansions are valid only if |x| < 1.

Application I Division = Multiple of x = M(x) Conclusion: The number by which division is to be made can be x or x2 or x3, but the number in the base is always expressed in the form of 1 + kx.

Class Exercise - 9 Solution : Which of the following expression is divisible by 1225? (a) (b) (c) (d) Solution : = 1225 k is divisible by 1225.

Application I Division is divisible by x – y for all positive integer n Proof: As each term is divisible by x – y, xn - yn is divisible by x – y

Application II Computation and Approximation Find 99993 exactly Solution : = 1000000000000 - 300000000 + 30000 -1 = 999700029999

Class Test

Class Exercise - 3 Solution : LHS = = RHS

Class Exercise - 4 Solution :

Class Exercise - 4 Solution : Compare the coefficient xn of both sides

Class Exercise - 5 Solution : LHS = = RHS

Class Exercise - 7 Find the coefficient of x4 in the expansion of Also find the coefficient of xr and find its expansion. Solution :

Solution Cont. Coefficient of x4 is 1.5 – 2.(–4) + 1.3 = 5 + 8 + 3 = 16 Coefficient of xr is

Class Exercise - 8 When x is so small that its square and higher powers may be neglected, find the value of Solution : As terms involving x2, x3, neglected

Solution Cont.

Thank you