mathematics. session binomial theorem session 2 session objectives

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Page 1: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Mathematics

Page 2: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Session

Binomial Theorem Session 2

Page 3: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Session Objectives

Page 4: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Session Objective

1. Properties of Binomial Coefficients

2. Binomial theorem for rational index

— General term

— Special cases

3. Application of binomial theorem

— Divisibility

— Computation and approximation

Page 5: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Properties of Binomial Coefficients

1. The sum of the binomial coefficients in the expansion of (1+x)n is 2n i.e.

n0 1 nC C ... C 2

where Cr = nCr

Proof:

n n n n 2 n n0 1 2 n1 x C C x C x ... C x

Put x = 1 both sides we get

n n n n n0 1 2 nC C C ...... C 2

For example 4C0 + 4C1 + 4C2 + 4C3 + 4C4

= 1 + 4 + 6 + 4 + 1 = 16 = 24

Page 6: Mathematics. Session Binomial Theorem Session 2 Session Objectives

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.

n–10 2 4 1 3 4C C C ... C C C ... 2

Proof: n n n n n0 1 n1 x C C x ... C x

Put x = –1 in above, we get

nn n n n0 1 2 n0 C C C ... C 1

n n n n n n0 2 4 1 3 5C C C ... C C C ...

As sum of all the coefficients is 2n

n n n0 22 C C ..... 2

n–10 2 4 1 3 4C C C ... C C C ... 2

n

rnr

r 0

C 1 0

Page 7: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Properties of Binomial Coefficients

3. n n 1 n 2r r 1 r 2

n n n 1C C . . C and so on

r r r 1

Proof:

n n–1

r r–2

n. n – 1 !n! nC C

r! n – r ! r. r – 1 ! n – r ! r

n–2r–2

n n – 1 n – 2 ! n n – 1. . C

r r – 1 r – 2 ! n – r ! r r – 1

For example: 4 3 22 1 0

4 4 3C . C . . C 6

2 2 1

Page 8: Mathematics. Session Binomial Theorem Session 2 Session Objectives

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

2 2 2 2 n–21 2 3 n1 C 2 C 3 C ... n C n n 1 2 .

Solution :

2 2 2 2 n–21 2 3 n1 C 2 C 3 C ... n C n n 1 2

n n2 n 2 n–1

r r–1r 1 r 1

nLHS r C r . . C

r

n

n–1r–1

r 1

n r C

n

n–1r–1

r 1

n r – 1 1 C

n n2 n n 1

r r 1r 1 r 1

r C n r 1 1 C

Page 9: Mathematics. Session Binomial Theorem Session 2 Session Objectives

n n

n–1 n–1r–1 r–1

r 1 r 1

n r – 1 C n C

n n

n–2 n–1r–2 r–1

r 2 r 1

n – 1n r – 1 C n C

r – 1

n n

n–2 n–1r–2 r–1

r 2 r 1

n n – 1 C n C

Solution Cont.

n n

n–2 n–1r–2 r–1

r 2 r 1

n n – 1 C n C

n–2 n–1 n–2n n – 1 2 n.2 n n 1 2

Page 10: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 2

Solution :

2 2 20 1 2 3

n 2n

C 2 C 3 C 4 C

... 1 n 1 C 0, n 0.

n n

r 2 r 2r r

r 0 r 0

LHS –1 r 1 C –1 r 2r 1 C

n

rr

r 0

–1 r r – 1 3r 1 C

n n nr r r

r r rr 0 r 0 r 0

1 r r 1 C 3 1 r C 1 C

Page 11: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Solution Cont.

n n nr r r

r r rr 0 r 0 r 0

1 r r 1 C 3 1 r C 1 C

n nr rn 1

r 1 rr 1 r 0

n3 1 r. . C 1 C

r

n r n 2

r 2r 2

n n 11 r r 1 . C

r r 1

n n

r rn–2 n–1r–2 r–1

r 2 r 1

n n – 1 –1 C 3n –1 C 0

= 0 + 0 + 0 = 0

Page 12: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Binomial theorem for Rational Index

Let n be a rational number and x be real number such that |x| < 1, then

n 2 3n n 1 n n 1 n 21 x 1 nx x x ...

2! 3!

rn n 1 n 2 ... n r 1... x ...

r!

Conditions of validity: if n is not a whole number

i) |x| < 1 ii) Number of terms is infinite

Page 13: Mathematics. Session Binomial Theorem Session 2 Session Objectives

General Term in (1+x)n, n Q

rr 1

n n 1 n 2 ... n r 1T x

r!

Expansion of (a + x)n for rational n

Case1: a

1x

n

n n ax a x 1

x

2 3n n n 1 n n 1 n 2a a a

x 1 n. ...x 2! x 3! x

Case2: x

1a

n

n n xx a a 1

a

2 3n n n – 1 n n – 1 n – 2x x x

a 1 n. ...a 2! a 3! a

Page 14: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 6

Solution :

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.

1

2 24 5x

–1

–1–1 2 22 22 5x4 – 5x 4 1–

4

22 2–1 –1

– 11 –1 –5x –5x2 2

(12 2 4 2! 4

32–1 –1 –1

– 1 – 2–5x2 2 2

...)3! 4

2 64

3

1 5x 75 1.3.5 125x1 x . ...

2 8 128 642 .3!

62 41 5 75 625 x

x x ...2 16 256 2048

Page 15: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Solution Cont.

Validity if2

25x 41 x

4 5

2 4 2x x

5 5

–2 2x

5 5

General termr2

r 1

–1 –1 –1– 1 ... – r 1

1 –5x2 2 2T

2 r! 4

r2

r

–1 –3 –5 ... –1– 2r 21 –5x.

2 42 r!

rr 2

r

–1 1.3.5... 2r – 11 –5x.

2 42 r!

r2r

r 1

1.3.5... 2r – 1 5x

4r! 2

Page 16: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Special Cases

n 2

3

n n 1i) 1 x 1 nx x

2!

n n 1 n 2x ...

3!

r rr 1

n n 1 ... n r – 1T –1 x

r!

n 2 3n n 1 n n 1 n 2ii) 1 x 1 nx x x ...

2! 3!

rr 1

n n 1 ... n r – 1T x

r!

Page 17: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Special Cases

n 2

3

n n 1iii) 1 x 1 nx x

2!

n n 1 n 2x ...

3!

r rr 1

n n – 1 ... n – r 1T –1 x

r!

Page 18: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Application

n n n 2 n n–1 n n1 2 n–1 n1 x 1 C x C x ... C x C x

n n n n n–11 2 n1 x – 1 x C C x ... C x

= Multiple of x = M(x)

n 21 x 1 nx M x

n 2 3n n – 11 x – 1– nx – x M x

2

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.

I Division

Page 19: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 9

Solution :

Which of the following expression is divisible by 1225?

(a) (b)

(c) (d)

2n6 – 35n – 1 2n6 – 35n 12n6 – 35n 2n6 35n – 1

21225 35

n2n n6 36 1 35

n 2 n n2 n1 35n C 35 ... C 35

2n 2 n n n n–22 3 n6 – 35n – 1 35 C C 35 ... C 35

= 1225 k2n6 – 35n – 1 is divisible by 1225.

Page 20: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Application

I Divisionn nx – y is divisible by x – y for all

positive integer n

Proof:

nn n nx – y x – y y – y

n n–1n n n n1 nx – y C x – y y ... C y – y

n n–1n n n–11 n–1x – y C x – y y ... C x – y y

As each term is divisible by x – y,

xn - yn is divisible by x – y

Page 21: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Application

II Computation and Approximation

Find 99993 exactly

Solution :

333 49999 10000 1 10 1

3 24 3 4 3 41 210 C 10 C 10 1

12 8 410 3.10 3.10 1

= 1000000000000 - 300000000 +30000 -1

= 999700029999

Page 22: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Test

Page 23: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 3

Solution :

31 2 n

0 1 2 n–1

n n 1CC C C2 3 ... n .

C C C C 2

n n

r r–1n! n – r 1n! n – r 1

C Cr! n – r ! r r – 1 ! n – r 1 ! r

nr r

nr–1r–1

C Cn – r 1

r CC

LHS = n n n

r

r–1r 1 r 1 r 1

C n – r 1r r. n – r 1

C r

n n 1 n n 1n 1 n –

2 2

= RHS

Page 24: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 4

Solution :

2 2 2 n0 1 n 2

2n ! 1.3.5... 2n – 1C C ... C 2

n!n!

2 2 20 1 nC C ... C

0 0 1 1 n nC .C C .C ... C .C

n no n 1 n–1 n 0 r n–rC .C C .C ... C .C as C C

n n0 1 n1 x C C x ... C x

n n–1n n–1 0C x C x ... C

n n n n0 1 n n 01 x . 1 x C C x ... C x C x ... C

Page 25: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 4

Solution :

2 2 2 n0 1 n 2

2n ! 1.3.5... 2n – 1C C ... C 2

n!n!

n n n n0 1 n n 01 x . 1 x C C x ... C x C x ... C

Compare the coefficient xn of both sides

2n 2 2 2n 0 1 nC C C ... C

2 2 2 2n0 1 n n

2n ! 1.2.3...2nC C ... C C

n! n! n! n!

n1.3.5.7...(2n – 1).2 .1.2...nn! n!

n1.3.5.7...(2n – 1).2n!

Page 26: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 5

Solution :

n 11 2 n

0C C C 2 – 1

C ...2 3 n 1 n 1

LHS = n n

r

r 0 r 0

C n!r 1 r! n – r ! r 1

n

r 0

n 1 !1n 1 r 1 ! n – r !

n

n 1 n 1r 1

r 0

1 1C 2 – 1

n 1 n 1

= RHS

Page 27: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 7

Solution :

Find the coefficient of x4 in the expansion

of Also find the coefficient of xr

and find its expansion.

21– x

.1 x

2

2 –21– x1– x 1 x

1 x –221– 2x x 1 x

2 2 32 2 1 2 3 41 x (1 2x x x

2! 3!

42 3 4 5x ...)

4!

2 3 41– 2x 3x – 4x 5x – ... r r

r 0

–1 r 1 x

Page 28: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Solution Cont.

2

2 2 3 41– x1– 2x x 1– 2x 3x – 4x 5x ...

1 x

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

Coefficient of xr is r r –1 r–2–1 r 1 – 2 –1 r –1 r – 1

r r–1 r 1 2r r – 1 –1 .4r

2

r r

r 1

1 x1 4 1 rx

1 x

Page 29: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Class Exercise - 8

Solution :

When x is so small that its square and higher powers may be neglected, find

the value of

–5

3

21 x 4 2x

3

4 x

1–5 –5

2

33 322

2 2 x1 x 4 2x 1 x 2 1

3 3 2

4 x x4 1

4

2 1 x1– 5 x 2 1 .

3 2 23 x

8 1 .2 4

As terms involving x2, x3, neglected

Page 30: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Solution Cont.

2 1 x1– 5 x 2 1 .

3 2 23 x

8 1 .2 4

10 x1– x 2

3 28 3x

–117x3 – 8 3x

6

11 17x 3x

3 18 6 8

1 17x 3x3 1

8 6 8

1 9x 17x3 – –

8 8 6

1 27 683 x

8 24

1 95x3 –

8 24

Page 31: Mathematics. Session Binomial Theorem Session 2 Session Objectives

Thank you