RESONANCE - Binomial Theorem

Part I

Part II

Part III

Part IV

Part - V.

Part - VI.

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DAILY LECTURE SHEET : ACADEMIC SESSION 2008-09

Title/Sub title of the Topic : ____________________________________________________________

Lecture No. of the particular topic : _____________________________________Cumulative Lecture No. ______

Contents to be discussed :

DPP No. _______ Home Assignments : ________________________________________________________

_________________________________________________________________________________________________

Theory Contents to be Covered in the Lecture : Assignments to be given after the Lecture :

Theory : 4 LectureDiscussion : 2 LectureBatch : R

Binomial Theorem for positive indexTheorem + basic properties

General term, middle term

Coefficient of xk in (ax + b)n

Num greatest coefficientNum greatest term

Remainder

1

Binomial Theorem

LECTURE 1

1. Expand the following binomials :(i) (x 3)5 [ Ans. x5 15x4 + 90x3 270x2 + 405x 243 ]

(ii) (2x y)5 [ Ans. 32x5 80x4y + 80x3y2 40x2y3 + 10xy4 y5 ]

2. Find

(i) 28th term of (5x + 8y)30 [ Ans. !3!27!30

(5x)3 . (8y)27]

(ii) 7th term of 9

x25

5x4

[Ans. 3x

10500]

3. Find middle term of (i) 142

2x1

& (ii)

93

6aa3

[Ans. (i) 16

429 x14, (ii) 8

189 a17, 16

21a19]

4. Find the term independent of x in 9

2

x31x

23

[Ans. 18

7]

D5. Find the coefficients of x32 & x17 in 15

34

x1x

[Ans. 1365, 1365]

6. Find the coefficient of x1 in (1 + 3x2 + x4) 8

x11

[Ans. 232]

7. Find the numerically greatest coefficient in the expansion of (3x 2)21 [ Ans. T9 ]

8. Find the algebracally greatest and least coefficient in the expansion of (2 5x)48[ Ans. T35 , T36 ]

9. Find the greatest term in the expansion of (7 5x)11 where x = 32

[Ans. T4 = 9440

78 53]

D10. Find numerically greatest term in the expansion of (3 5x)15 when x = 51

. [ Ans. T3, T4 ]

11. If n is a positive integer, show that(i) 9n + 7 is divisible by 8.D(ii) 32n+1 + 2n+2 is divisible by 7D(iii) 62n 35n 1 is divisible by 1225.

12. What is the remainder when 599 is divided by 13 [Ans. 8]

D13. What is the remainder when 7103 is divided by 25. [Ans. 18]

Part I

Part II

Part III

Part IV

Part - V.

Part - VI.


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Sum of series involves binomial coefficientupper index is constant(i) By proper substitution

(ii) kt = knt(iii) By reduction r nCr = n n1Cr1(iv) Calculus

rC , rCr , rC)1( , rC1r1

,

r2Cr ...

2

(A + B )n

Binomial Theorem

LECTURE 2

1. If n is +ve integer, prove that the integral part of (7 + 4 3 )n is an odd number..

D2. If n is +ve integer, prove that the integeral part of (7 + 2 5 )2n + 1 is an even number..

D3. If (7 + 4 3 )n = , where is a +ve integer and is a proper fraction, then prove that

(1) () = 1

4. Show that the integer just above ( 3 +1)2n is divisible by 2n + 1 for all n N,

5. If (1 + x)n = C0 + C1x + C2x2 + .........+ Cnxn, then show that

(i) C0 + C1 + C2 + ..........+Cn = 2n

(ii) C0 + C2 + C4 + .......... = 2n1

D(iii) C1 + C3 + C5 + .......... = 2n1

(iv) C0 + 3C1 + 32C2 + ........ = 4n.

6. If (1 + x)n = C0 + C1x + C2x2 + .........+ Cnxn, then show that

(i) C1 + 2C2 + 3.C3 + ............. + nCn = n.2n1

(ii) C0 + 2C1 + 3.C2 + ............. + (n+1) . Cn = 2n1(n+2)

(iii) C0 + 3C1 + 5C2 + ............. + (2n+1) . Cn = 2n(n+1)

D(iv) a.C0 + (a b).C1 _ (a 2b)C2 + .........+(a nb).Cn = 2n1 (2a nb)

(v) C0 22.C1 + 32 C2 + .............+ (1)n (n+1)2 .Cn = 0, n > 2.

D(vi) a2.C0 (a 1)2.C1 + (a 2)2.C2 (a 3)2.C3 + ..........+ (1)n (a n)2.Cn = 0.

(vii)1n

121n

C..........3

C2

CC1n

n210

(viii) 1n1..........

4C

3C

2CC 3210

(ix)

n

0r

rr

2n1

)2r)(1r(C.)1(

.

(x)

n

0r 1r2r

Cr = 1n1)3n(2n

.

Part I

Part II

Part III

Part IV

Part - V.

Part - VI.


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3

Problem of double sigma

Product of binomial coefficient

Upper index is variable(i) nCr + nCr + 1 = n+1Cr+1(ii) Using binomial theorem for negative index

Binomial Theorem

LECTURE 3

1. (i) nCn + n+1Cn + n+2Cn + ........+ n+kCn = n+kCn+1 .

(ii) mC1 + m+1C2 + m+2C3 + ....... + m+n1Cn = nC1 + n+1C2 + n+2C2 + .......+n+m1Cm.

2. Prove that If (1 + x)n = C0 + C1x + C2x2 +..............+ Cnxn,

(i) C02 + C12 + ..... + Cn2 = )!n(!)n2(

(ii) C0C1 + C1C2 + ...... + Cn1 Cn =

)!1n()!1n(!n2

(iii) C0Cn + C1 . Cn1 + .............. + Cn C0 = 2)!n()!n2(

D(iv)

n

0r3rr 3)!-(n3)!(n

! n2CC .

(v) mCr.nC0 + mCr1.nC1 + mCr2.nC2 + .......+mC1.nCr1 + mC0.nCr = m+nCr ,

where m,n,r are positive integers and r < m, r < n.

D(vi) C0.2nCn C1.2n2Cn + C2. 2n4Cn .......... = 2n, where Cr stands for nCr.

3. Evaluate :

n

0m

m

0pp

mm

n C.C Ans. 3n

4. If (1 + x)n = C0 + C1x + C2x2 + ..........+ Cnxn, prove that

(i) injn CC)ji( = n.22nD(ii) (Ci + Cj)2 = (n 1).2nCn + 22n 0 i < j n,

D5. If (1 + x)n = C0 + C1x + C2x2 + ............+ Cnxn, show that

i.j.CiCj=n2 { 1n2n23n2 C.

212

} 0 i < j n.

D6. Prove that If (1 + x)n = C0 + C1x + C2x2 +..............+ Cnxn,

(i) 22

n2

22

120 ]! 1)[(n

! )1n2(1n

C.......3

C2

CC

(ii) C12 2.C22 + 3.C32 ........ 2n.C22n = (1)n1.n.Cn where Cr stands for 2nCr.

(iii)

n

0rn

2n222r C.nC)rn(r , where Cr stands for nCr.

(iv) C0 C1 + C2 ............+ (1)m1.Cm1 = (1)m1 ! )1m()1mn).....(2n)(1n(

where Cr stands for nCr.

D7. If (1 + x)n = C0 + C1x + C2x2 + ..........+ Cnxn, show that

n

0r r

2

ji C1

2n

Cj

Ci

, 0 i < j n.

Part I

Part II

Part III

Part IV

Part - V.

Part - VI.


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Binomial theorem for negative and fractional index

Multinomial theorem

Binomial Theorem

4

LECTURE 4

1. Write the expansion of(i) (1 x)1 (ii) (1 x)2 (iii) (1 x)n

2. Find the general term in the expansion of (23x2)2/3 [Ans. r32

2!r)1r3......(8.5.22

x2r]

D3. Find the no. of term in the expansion of (a + b + c + d + e + f)n. [Ans. n+5cn]

D4. Find the coeff. of a3.b4.c7 in the expansion of (bc + ca + ab)8. [Ans. 280]

5. Find the coefficient of x3 y4 z2 in the expansion of (2x 3y + 4z)9. Ans. !2!4!3!9

23 34 42

6. Find the coefficient of x4 in (1 + x 2x2)7 Ans. 14

D7. The sum of rational terms in

1063 532 is ............... [Ans. 12632]

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RESONANCE - Binomial Theorem