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September 25, 2017 | Author: suar90 | Category: Integer, Discrete Mathematics, Algebra, Mathematical Objects, Numbers
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BANSALCLASSES TARGET IIT JEE 2008

M A T H E M A T I C S ACME

BINOMIAL CONTENTS KEY- CONCEPTS EXERCISE - I(A) EXERCISE - I(B) EXERCISE - II EXERCISE - III(A) EXERCISE - III(B) EXERCISE - IV ANSWER - KEY

KEY CONCEPTS BINOMIAL EXPONENTIAL & LOGARITHMIC SERIES 1.

BINOMIAL THEOREM : The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as BINOMIAL THEOREM . If x , y  R and n  N, then ; n

(x +

y)n

=

nC 0

xn

+

nC 1

xn1

y+

nC xn2y2 2

+ ..... +

nC xnryr r

+ ..... +

nC yn n

=

 nCr xn – r yr. r0

This theorem can be proved by Induction . OBSERVATIONS : (i) The number of terms in the expansion is (n + 1) i.e. one or more than the index . (ii) The sum of the indices of x & y in each term is n . (iii) The binomial coefficients of the terms nC0 , nC1 .... equidistant from the beginning and the end are equal. 2. (i) (iii)

IMPORTANT TERMS IN THE BINOMIAL EXPANSION ARE : General term (ii) Middle term Term independent of x & (iv) Numerically greatest term

(i)

The general term or the (r + 1)th term in the expansion of (x + y)n is given by ; Tr+1 = nCr xnr . yr

(ii)

The middle term(s) is the expansion of (x + y)n is (are) : (a) If n is even , there is only one middle term which is given by ; T(n+2)/2 = nCn/2 . xn/2 . yn/2 (b) If n is odd , there are two middle terms which are : T(n+1)/2 & T[(n+1)/2]+1

(iii)

Term independent of x contains no x ; Hence find the value of r for which the exponent of x is zero.

(iv)

To find the Numerically greatest term is the expansion of (1 + x)n , n  N find Tr 1 Tr

n



n

Cr x r

C r 1x

inequality

r 1



n  r 1 x . Put the absolute value of x & find the value of r Consistent with the r

Tr 1

> 1. Tr Note that the Numerically greatest term in the expansion of (1  x)n , x > 0 , n  N is the same as the greatest term in (1 + x)n . 3.

If



A B



n

= I + f, where I & n are positive integers, n being odd and 0 < f < 1, then

(I + f) . f = Kn where A  B2 = K > 0 & A  B < 1. If n is an even integer, then (I + f) (1  f) = Kn. 4.

BINOMIAL COEFFICIENTS :

(i)

C0 + C1 + C2 + ....... + Cn = 2n

(ii)

C0 + C2 + C4 + ....... = C1 + C3 + C5 + ....... = 2n1

(iii)

C0² + C1² + C2² + .... + Cn² = 2nCn =

(iv)

C0.Cr + C1.Cr+1 + C2.Cr+2 + ... + Cnr.Cn =

Bansal Classes

(2 n) ! n! n!

( 2n )! ( n  r ) ( n  r )!

Binomial

[2]

REMEMBER : (i) (2n)! = 2n . n! [1. 3. 5 ...... (2n  1)] 5.

BINOMIAL THEOREM FOR NEGATIVE OR FRACTIONAL INDICES : If n  Q , then (1 + x)n = 1  n x 

n ( n  1) 2 n ( n  1) ( n  2) 3 x  x ...... Provided | x | < 1. 2! 3!

Note : (i) When the index n is a positive integer the number of terms in the expansion of (1 + x)n is finite i.e. (n + 1) & the coefficient of successive terms are : nC , nC , nC , nC ..... nC 0 1 2 3 n (ii)

When the index is other than a positive integer such as negative integer or fraction, the number of terms in the expansion of (1 + x)n is infinite and the symbol nCr cannot be used to denote the Coefficient of the general term .

(iii)

Following expansion should be remembered (x < 1). (a) (1 + x)1 = 1  x + x2  x3 + x4  ....  (b) (1  x)1 = 1 + x + x2 + x3 + x4 + ....  (c) (1 + x)2 = 1  2x + 3x2  4x3 + ....  (d) (1  x)2 = 1 + 2x + 3x2 + 4x3 + ..... 

(iv)

The expansions in ascending powers of x are only valid if x is ‘small’. If x is large i.e. | x | > 1 then 1 we may find it convinient to expand in powers of , which then will be small. x APPROXIMATIONS :

6.

n ( n  1) n ( n  1) ( n  2) 3 x² + x ..... 1. 2 1.2.3 If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its squares and higher powers may be neglected then (1 + x)n = 1 + nx, approximately. This is an approximate value of (1 + x)n.

(1 + x)n = 1 + nx +

7.

EXPONENTIAL SERIES :

(i)

ex = 1 +

1  x x2 x3    ....... ; where x may be any real or complex & e = Limit 1   n  n 1! 2! 3!

(ii)

ax = 1 +

x x2 2 x3 3 ln a  ln a  ln a  ....... where a > 0 1! 2! 3!

n

Note : (a) (b)

(c) (d) (e)

1 1 1   ....... 1! 2! 3! e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is 2.7182818284.

e=1+

 1 1 1  e + e1 = 2 1     .......   2! 4! 6!   1 1 1  e  e1 = 2 1     .......   3! 5! 7!  Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier, their inventor. They are also called Natural Logarithm.

Bansal Classes

Binomial

[3]

8.

LOGARITHMIC SERIES :

(i)

ln (1+ x) = x 

(ii)

ln (1 x) =  x 

(iii)

(1  x ) ln =2 (1 x )

REMEMBER :

(a) (c)

x 2 x3 x 4    .......  where 1 < x  1 2 3 4 x 2 x3 x 4    .......  where 1  x < 1 2 3 4

3 5    x  x  x  ......  x < 1   3 5  

1 1 1   +...  = ln 2 2 3 4 ln2 = 0.693

1

(b)

eln x = x

(d)

ln10 = 2.303

EXERCISE–I (A) 11

11

 1  (ii) x7 in  ax  2  bx   (iii) Find the relation between a & b , so that these coefficients are equal.

Q.1

 1  Find the coefficients : (i) x7 in  a x 2   bx  

Q.2

If the coefficients of (2r + 4)th , (r  2)th terms in the expansion of (1 + x)18 are equal , find r.

Q.3

If the coefficients of the rth, (r + 1)th & (r + 2)th terms in the expansion of (1 + x)14 are in AP, find r. 10

8

Q.4

 x 3  2 Find the term independent of x in the expansion of (a)   3 2x 

Q.5

Find the sum of the series

Q.6

If the coefficients of 2nd , 3rd & 4th terms in the expansion of (1 + x)2n are in AP, show that 2n²  9n + 7 = 0.

Q.7

Given that (1 + x + x²)n = a0 + a1x + a2x² + .... + a2nx2n , find the values of : (i) a0 + a1 + a2 + ..... + a2n ; (ii) a0  a1 + a2  a3 ..... + a2n ; (iii) a02 a12 + a22  a32 + ..... + a2n2

Q.8

If a, b, c & d are the coefficients of any four consecutive terms in the expansion of (1 + x)n, n  N,

 1 1/ 3 1 / 5  (b)  x  x  2 

 1 3r  7 r 15r r n    4r  .....up to m terms  (  1 ) . C   r r 2r 3r 2 2 2 2  r0 n

a c 2b prove that a  b  cd  bc .

Q.10

 2 log Find the value of x for which the fourth term in the expansion,  5 5 5   Prove that : n1Cr + n2Cr + n3Cr + .... + rCr = nCr+1.

Q.11

(a) Which is larger : (9950 + 10050) or (101)50.

Q.9

(b) Show that

Bansal Classes

2n–2C n–2

+ 2.2n–2Cn–1 + 2n–2Cn >

8

x

4  44

1



log 5 2 x 1 7 3

5

  is 336.  

4n , nN , n  2 n 1

Binomial

[4]

Q.12

7  In the expansion of 1  x    x

11

find the term not containing x.

Q.13 Show that coefficient of x5 in the expansion of (1 + x²)5 . (1 + x)4 is 60. Q.14 Find the coefficient of x4 in the expansion of : (i) (1 + x + x2 + x3)11 (ii) (2  x + 3x2)6 Q.15 Find numerically the greatest term in the expansion of : (i) (2 + 3x)9 when x =

3 2

(ii) (3  5x)15 when x =

Q.16 Given sn= 1 + q + q² + ..... + qn & n+1C 1

prove that

Sn = 1 +

1 5

 q  1

q 1 2

2

+  2 

 q  1

n

+ .... +  2  , q  1,

+ n+1C2.s1 + n+1C3.s2 +....+ n+1Cn+1.sn = 2n . Sn .

Q.17 Prove that the ratio of the coefficient of x10 in (1  x²)10 & the term independent of x in 2  x    x

10

is 1 : 32 .

Q.18 Find the term independent of x in the expansion of

 3 x2 1  (1 + x + 2x3)    2

3x 

9

.

Q.19 In the expansion of the expression (x + a)15 , if the eleventh term is the geometric mean of the eighth and twelfth terms , which term in the expansion is the greatest ? n4

Q.20 Let (1+x²)² . (1+x)n =

a

K

. x K . If a1 , a2 & a3 are in AP, find n.

K 0

Q.21 If the coefficient of ar–1 , ar , ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n (4r + 1) + 4r2 – 2 = 0. Q.22 If

nJ r

(1  x n )(1  x n 1 )(1  x n 2 )..................(1  x n  r 1 ) = , prove that nJn – r = nJr.. (1  x )(1  x 2 )(1  x 3 )..................(1  x r ) n

Q.23 Prove that

 n C K sin Kx . cos(n  K )x  2n 1 sin nx . K 0

Q.24 The expressions 1 + x, 1+x + x2, 1 + x + x2 + x3,............. 1 + x + x2 +........... + xn are multiplied together and the terms of the product thus obtained are arranged in increasing powers of x in the form of a0 + a1x + a2x2 +................., then, (a) how many terms are there in the product. (b) show that the coefficients of the terms in the product, equidistant from the beginning and end are equal. (c)

show that the sum of the odd coefficients = the sum of the even coefficients =

Q.25 Find the coeff. of

2n

(a) (b) (c)

(n  1)! 2

x6 in the expansion of (ax² + bx + c)9 . x2 y3 z4 in the expansion of (ax  by + cz)9 . a2 b3 c4 d in the expansion of (a – b – c + d)10.

2n

r r Q.26 If  a r (x2)  b r ( x3) & ak = 1 for all k  n, then show that bn = 2n+1Cn+1. r0

Bansal Classes

r 0

Binomial

[5]

i k 1

Q.27 If Pk (x) =

 xi

n

1 x   2 

 n C k Pk ( x)  2n 1 ·Pn 

then prove that,

i 0

k 1

Q.28 Find the coefficient of xr in the expression of : (x + 3)n1 + (x + 3)n2 (x + 2) + (x + 3)n3 (x + 2)2 + ..... + (x + 2)n1 x 5

n

2 5

Q.29(a) Find the index n of the binomial    if the 9th term of the expansion has numerically the   greatest coefficient (n  N) . (b) For which positive values of x is the fourth term in the expansion of (5 + 3x)10 is the greatest.

(72)!

Q.30 Prove that

36!2

 1 is divisible by 73.

Q.31 If the 3rd, 4th, 5th & 6th terms in the expansion of (x + y)n be respectively a , b , c & d then prove that

b 2  ac 5a  . c 2  bd 3c Q.32 Find x for which the (k + 1)th term of the expansion of (x + y)n is the greatest if x + y = 1 and x > 0, y > 0. *Q.33 If x is so small that its square and higher powers may be neglected, prove that : (1  3 x )1/2  (1  x ) 5 / 3  41  1  x 1/ 2  24  (4  x)

(i)

(ii)

1   1 

x 2

1/ 3

    1  

3x 7

 1

1/3

3x 5

7x 3

5

1/7

= 1 1

 x   x  x 10 7 127 84

*Q.34 (a)

If x =

1 1. 3 1. 3. 5 1. 3. 5 . 7     ........  then prove that x2 + 2x – 2 = 0. 3 3. 6 3. 6 . 9 3 . 6 . 9 .12

(b)

If y =

2 1. 3  5 2!

2

1. 3 . 5  2     5 3!

1 12

or

3

 2    ........ then find the value of y² + 2y..  5 1/ n

( n  1) p  ( n  1)q  p    *Q.35 If p = q nearly and n >1, show that ( n  1) p  ( n  1)q  q 

(*)

.

Only for CBSE. Not in the syllabus of IIT JEE.

EXERCISE–I (B) Q.1

Show that the integral part in each of the following is odd. n  N



(A) 5  2 6 Q.2



n



(B) 8  3 7



n



(C) 6  35



n

Show that the integral part in each of the following is even. n N



(A) 3 3  5





2n 1





(B) 5 5  11

2n  1



Q.3

If 7  4 3 n = p+ where n & p are positive integers and  is a proper fraction show that (1  ) (p + ) = 1.

Q.4

If x denotes 2  3 , n N & [x] the integral part of x then find the value of : x  x² + x[x].

Bansal Classes





n

Binomial

[6]

Q.5



If P = 8  3 7



n

and f = P  [P], where [ ] denotes greatest integer function.

Prove that : P (1  f) = 1 (n  N)





2n  1

= N & F be the fractional part of N, prove that NF = 202n+1 (n  N)

Q.6

If 6 6  14

Q.7

  Prove that if p is a prime number greater than 2, then the difference  2  5   2p+1 is divisible by  





p

p, where [ ] denotes greatest integer.





2n

contains 2n+1 as factor (n  N)

Q.8

Prove that the integer next above

Q.9

Let I denotes the integral part & F the proper fractional part of 3  5

3 1





n

where n  N and if 

denotes the rational part and  the irrational part of the same, show that 1 1 (I + 1) and  = (I + 2 F  1). 2 2

=

2n

Q.10 Prove that

Cn

n 1

is an integer,  n  N.

EXERCISE–II (NOT IN THE SYLLABUS OF IIT-JEE) PROBLEMS ON EXPONENTIAL & LOGARITHMIC SERIES For Q.1 TO Q.15, Prove That : 2

2

Q.1

    1 1 1 1 1 1    ......   1     ...... 1  2! 4 ! 6 ! 3! 5! 7!    

Q.2

 1  e1 1 1 1 1 1     ......      ...... e  1  2! 4 ! 6 !   1! 3! 5! 

Q.3

   e2  1  1 1 1 1 1 1     ......   1     ...... 2! 4! 6! e 2  1  1! 3! 5!   

Q.4

1+

Q.5 Q.6 Q.7

=1

1 2 1 2  3 1 2  3 4  3    .......     e  2 2! 3! 4!

1 1 1 1    ........  = 1. 3 1. 2 . 3. 5 1. 2 . 3. 4 . 5 . 7 e

1+ 23

1  2 1  2  2 2 1  2  22  23    ........ 2! 3! 4! 33

43

1 + 2!  3!  4!  ........ = 5e

Q 8.

= e² e 2 3 6 11 18      ........ 1! 2! 3! 4 ! 5!

Q.9

1 1 1    ........  = 1  loge 2 2.3 4.5 6 .7

Q.11

1 1 1 1 1 1 1    ... =     .... = ln 2 1. 2 3. 4 5 . 6 2 1. 2 . 3 3. 4 . 5 5 . 6 . 7

Q.12

1 1 1 1    2 2 . 22 3. 23 4 .24

Bansal Classes

+..... = ln3  ln2

Q 10. 1 +

Q 13.

Binomial

= 3 (e  1)

1 1 1    ........  = loge3 2 4 3. 2 5.2 7 . 26

1 1 1 1    3 3 . 33 5 . 35 7 . 37

 1

+..... =   ln 2  2

[7]

Q.14

1  1 1 1  1 1  1 1 1      2  2    3  3   .......  l n 2 2  2 3 4  2 3  62 3 

2 3 4 y2 y3 y4   Q.15 If y = x  x  x  x +..... where | x | < 1, then prove that x = y + +......

2

3

2!

4

3!

4!

EXERCISE–III (A) If C0 , C1 , C2 , ..... , Cn are the combinatorial coefficients in the expansion of (1 + x)n, n  N , then prove the following :

2 n !

Q.1

C0² + C1² + C2² +.....+ Cn² =

Q.2

C0 C1 + C1 C2 + C2 C3 +....+Cn1 Cn =

Q.3

C1 + 2C2 + 3C3 +.....+ n . Cn = n . 2n1

Q.4

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

Q.5

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

Q.6

n (C0+C1)(C1+C2)(C2+C3) ..... (Cn1+Cn) = C 0 . C1 . C 2 .... C n 1 (n  1)

n! n !

2 n ! ( n  1)! ( n  1)!

n!

Q.7

C1 2 C2 3 C3 n .Cn n ( n  1)    .......   C0 C1 C2 C n 1 2

Q.9

2 . Co +

Q 8.

C

C1 C 2 C 2 n 1  1   ......  n  2 3 n 1 n 1

22 . C1 23 . C 2 24 .C3 2n 1 .C n 3n  1  1    ......  2 3 4 n 1 n 1

Q.10 CoCr + C1Cr+1 + C2Cr+2 + .... + Cnr Cn = Q.11

C0 +

C

C

2 n! (n  r)! (n  r)!

1

n 1 2 n Co  2  3  ......  ( 1) n  1  n  1

Q.12 Co  C1 + C2  C3 + .... + (1)r . Cr =

(  1) r ( n  1)! r ! . ( n  r  1)!

Q.13 Co  2C1 + 3C2  4C3 + .... + (1)n (n+1) Cn = 0 Q.14 Co²  C1² + C2²  C3² + ...... + (1)n Cn² = 0 or (1)n/2 Cn/2 according as n is odd or even. Q.15 If n is an integer greater than 1 , show that ; a  nC1(a1) + nC2(a2)  ..... + (1)n (a  n) = 0 Q.16 (n1)² . C1 + (n3)² . C3 + (n5)² . C5 +..... = n (n + 1)2n3 Q.17 1 . Co² + 3 . C1² + 5 . C2² + ..... + (2n+1) Cn² =

(n  1) (2 n)! n! n !

Q.18 If a0 , a1 , a2 , ..... be the coefficients in the expansion of (1 + x + x²)n in ascending powers of x , then prove that : (i) a0 a1  a1 a2 + a2 a3  .... = 0 . (ii) a0a2  a1a3 + a2a4  ..... + a2n  2 a2n = an + 1 or an–1. (iii)

E1 = E2 = E3 = 3n1 ; where E1= a0 + a3 + a6 + ..... ; E2 = a1 + a4 + a7 + ..... & E3 = a2 + a5 + a8 + ..... n2

Q.19 Prove that :



r0

n



Cr . nCr  2 =

(2 n)! (n  2)! ( n  2)!

Q.20 If (1+x)n = C0 + C1x + C2x² + .... + Cn xn , then show that the sum of the products of the C i ’ s taken two at a time , represented by

Bansal Classes

  Ci C j 2 n! is equal to 22n1  . 0 i j n 2 ( n !) 2

Binomial

[8]

C n  2n 1 

n 1 2

Q.21

C1 

Q.22

C1  C2  C3  ......  C n  n 2 n  1

C2 

C 3  ...... 



1/ 2



for n  2.

EXERCISE–III (B) Q.1

If (1+x)15 = C0 + C1. x + C2. x2 + .... + C15. x15, then find the value of : C2 + 2C3 + 3C4 + .... + 14C15

Q.2

If (1 + x + x² + ... + xp)n = a0 + a1x + a2x²+...+anp. xnp , then find the value of : a1 + 2a2 + 3a3 + .... + np . anp

Q.3

1². C0 + 2². C1 + 3². C2 + 4². C3 + .... + (n+1)² Cn = 2n2 (n+1) (n+4) . n

Q.4

r

. C r  n (n  1) 2 n  2

2

r0 n

Q.5

Given p+q = 1 , show that

r

2

. n C r . p r . q n  r  n p (n  1) p  1

r0 n

Q.6

Show that

 C  2 r  n r

r0

(1 +

2

 n . 2n where C denotes the combinatorial coeff. in the expansion of r

x)n. C1 C C Cn (1  x) n  1  1 x  2 x 2  3 x 3  .......  . xn = (n  1) x 2 3 4 n1

Q.7

C0 +

Q.8

Prove that , 2 . C0 +

Q.9

If (1+x)n =

22 23 211 311  1 . C1  . C 2  ......  . C 10  2 3 11 11

n

C

r

. xr

then prove that ;

r0

22 . C 0 2 3 . C1 24 .C2 2n  2 .C n 3n  2  2n  5    ......   1. 2 2.3 3. 4 (n  1) (n  2) (n  1) (n  2)

Q.10

C0 C C 2n  2  4  ........  1 3 5 n 1

Q.11

C 0 C1 C 2 C 3 Cn 4 n . n!     ........(1) n  1 5 9 13 4 n  1 1. 5 . 9 .13..... (4 n  3) (4 n  1)

Q.12

C0 C C C C 1 n .2n 1  1  2  3  ........  n  2 3 4 5 n  2 (n  1) (n  2)

Q.13

C0 C C C C 1  1  2  3  .......  ( 1) n . n  2 3 4 5 n  2 (n  1) (n  2)

Q.14

1 1 1 1 C1 C 2 C 3 C 4 C     .......  ( 1) n  1 . n = 1     .......  1 2 3 4 n 2 3 4 n

Q.15 If (1+x)n = C0 + C1x + C2x² + ..... + Cn xn , then show that : C1(1x) 

C2 2

(1x)² +

Bansal Classes

C3 3

1 n

(1x)3 ....+ (1)n1 (1x)n = (1x) +

Binomial

1 2

(1x²) +

1 3

1 n

(1x3) +......+ (1xn)

[9]

1n 2 3 4 C1 nC2+ nC3 nC4 + 2 3 4 5

Q.16 Prove that ,

n

Q.17 If n  N ; show that

..... +

1 ( 1) n  1 n n . Cn= n 1 n 1

n n n C0 C1 C2 Cn n!    ......  (1) n  x x 1 x  2 x  n x (x  1) (x  2) .... (x  n )

Q.18 Prove that , (2nC1)²+ 2 . (2nC2)² + 3 . (2nC3)² + ... + 2n . (2nC2n)² =

(4 n  1)!

(2 n  1) !2

2n

Q.19

If (1 + x + x2)n

=

 ar xr , n  N , then prove that

r0

(r + 1) ar + 1 = (n  r) ar + (2n  r + 1) ar1. Q.20 Prove that the sum to (n + 1) terms of

( 0 < r < 2n)

C0 C1 C2    ......... equals n (n  1) (n  1) (n  2) (n  2) (n  3)

1



xn1. (1 x)n+1 . dx & evaluate the integral.

0

EXERCISE–IV Q.1

The sum of the rational terms in the expansion of

Q.2

If an =

n

n

1

r

 n , then  n r0 C C r 0

r

(A) (n1)an



2  31/ 5

10



is ___ .

[JEE ’97, 2]

equals

[JEE’98, 2]

r

(B) n an

(C) n an / 2

(D) None of these

3 5 9 15 23      ........  1! 2! 3! 4! 5!

Q.3

Find the sum of the series

[REE ’98, 6]

Q.4

If in the expansion of (1 + x)m (1  x)n, the co-efficients of x and x2 are 3 and  6 respectively, then m is : [JEE '99, 2 (Out of 200)] (A) 6 (B) 9 (C) 12 (D) 24  n   n   +   =  r  1  r  2

 n  r

Q.5(i) For 2  r  n ,   + 2   n  1   r  1

(A) 

 n  1   r  1

(B) 2 

 n  2   r 

(C) 2 

 n  2   r 

(D) 

(ii) In the binomial expansion of (a  b)n , n  5 , the sum of the 5th and 6th terms is zero . Then

a equals: b

[ JEE '2000 (Screening) , 1 + 1 ] (A)

n5 6

Bansal Classes

(B)

n4 5

(C)

5 n4

Binomial

(D)

6 n5

[10]

Q.6

 n  = nCm . Prove that  m

For any positive integers m , n (with n  m) , let 

 n  1  n  2  m  n  n  1  +   + ........ +   =    +    m   m   m  m  m  1

Hence or otherwise prove that ,  n  1  n  2  n  m  n2  + 3   + ........ + (n  m + 1)   =    + 2   . m m      m  m  m  2

[ JEE '2000 (Mains), 6 ] Q.7 Q.8

Find the largest co-efficient in the expansion of (1 + x) n , given that the sum of co-efficients of the terms in its expansion is 4096 . [ REE '2000 (Mains) ] a In the binomial expansion of (a – b)n, n > 5, the sum of the 5th and 6th terms is zero. Then equals b (A)

Q.9

n5 6

(B)

n4 5

(C)

5 n4

6 n5 [ JEE '2001 (Screening), 3] (D)

Find the coeffcient of x49 in the polynomial

[ REE '2001 (Mains) , 3 ]

  C   2 C   2 C  2 C   x  1   x  2  2   x  3  3  .................  x  50  50  C0   C1   C2  C 49   

where Cr = 50Cr .

m

Q.10 The sum 

  , 10 20 i m i

(where

i 0

(A) 5

  = 0 if P < q ) is maximum when m is p q

(B) 10

Q.11(a) Coefficient of t24 (A) 12C6 + 2

(C) 15

(D) 20 [ JEE '2002 (Screening) , 3 ] 2 12 12 24 in the expansion of (1+ t ) (1 + t ) (1 + t ) is (B) 12C6 + 1 (C) 12C6 (D) none [JEE 2003, Screening 3 out of 60]

 n  n   n  n  1   n  n  2   n  n  K   n  (b) Prove that : 2K .  0  K  – 2K–1 1  K  1 + 2K–2  2  K  2  ...... (–1)K  K  0    K  .               [JEE 2003, Mains-2 out of 60] Q.12 n–1Cr = (K2 – 3).nCr+1, if K  (A) [– 3 ,

3]

 30  Q.13 The value of  0    30 (A)  10   

Bansal Classes

(B) (–, – 2)

 30   30   10  –  1       30  (B)  15   

(C) (2, )

 30   30   11  +  2     

 30   12  ........ +  

 60  (C)  30   

Binomial

(D) ( 3 , 2] [JEE 2004 (Screening)]  30   30   20   30  is, where   

n n  r  = Cr.  

 31 (D) 10    [JEE 2005 (Screening)]

[11]

ANSWER KEY EXERCISE–I (A) Q 1. Q 5.

(i)

11C 5

2

1

mn

a6 a5 (ii) 11C6 6 (iii) ab = 1 Q 2. r = 6 5 b b



Q 7. (i) 3n (ii) 1, (iii) an

2  1 2  n

mn

Q 3. r = 5 or 9 Q 4. (a)

Q 9. x = 0 or 1

5 12

(b) T6 =7

Q 10. x = 0 or 2 5

Q 11.

(a)10150 (Prove

that

10150 

9950 =

10050 +

some +ive qty) Q 12. 1 +

 11C2k . 2kCk 7k

k1

Q 14. (i) 990 (ii) 3660 Q.19 T8

Q 15. (i) T7 =

7.313 2

(ii) 455 x 312

Q 18.

17 54

Q.20 n = 2 or 3 or 4

n2  n  2 2 6 3 Q 25. (a) 84b c + 630ab4c4 + 756a2b2c5 + 84a3c6

Q.24 (a)

Q 28. nCr (3nr  2nr)

Q 29. (a) n = 12 (b)

; (b) 1260 . a2b3c4 ; (c) 12600 20 5
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