pages from final algebra consolidated

1. Binomial Expression An algebraic expression consisting of only two terms is called a binomial expression. For example, expressions such as x + a, 4x + 3y, ( 2x 4 __ y ) are allbinomial expressions.

2. Binomial Theorem This theorem gives a formula by which any power of a binomial expression can be expanded was fi rst given by Isaac Newton.

2.1 Binomial Theorem for Positive Integral Index If x and a are real numbers, then for all n N,

(x + a)n = nC0 xn a0 + nC1 x

n 1 a1 + nC2 xn 2 a2 + .....

.... + nCn 1 x1 an 1 + nCn x0 a

n (1)

OR

(x + a)n = (nC0 xn0 a0 + nC2 x

n2a2 + nC4xn4a4 + .....)

+ (nC1xn 1a1 +

nC3xn 3 a3

+ nC5xn 5 a5 + .......)

= A + B (say) = Sum of odd terms + Sum of even

terms Here nC0,

nC1, nC2, ......,

nCn are called binomial coefficients. These are generally denoted by C0, C1, C2, .......,Cn.

Note 1: The positive integer n is called the index of the binomial.

Note 2: The number of terms in the expansion of (x + a)n is n + 1, i.e., one more than the index n.

Note 3: In the expansion of (x + a)n, the power of x goes on decreasing by 1 and that of a goes on increasing by 1 so that the sum of powers of x and a in any term is n.

Note 4: The binomial coeffi cients of the terms are equidistant from the beginning i.e., nCr =

nCn r.

Note 5: Binomial coeffi cient of (r + 1)th term is = nCr i.e. the number of terms is one more than the value of r.

2.2 General Term in the Expansion of (x + a)n In the binomial expansion of (x + a)n the (r + 1)th term from the beginning is usually called the general term and it is denoted by Tr + 1, i.e., Tr + 1 =

nCr xn r ar = nCr (first term)

n r. (second term)r

It is obvious to note that the binomial coeffi cient of the general term

i.e., (r + 1)th term = nCr . 2.3 General term in the expansion of

(a + x)n is Tr + 1 =

nCr an r xr.

3. Special Cases (i) Replacing a by a, in (1), we get

L E C T U R E

Binomial Expansion

BASIC CONCEPTS

1

A.4 Binomial Expansion

(x a)n = nC0xna0 nC1x

n 1a1 + nC2xn 2a2 ...

+ ( 1)n nCnx0an. (2)

(ii) Replacing x by 1 and a by x, we get (1 + x)n = nC0 +

nC1x + nC2x

2 + ... + nCnx

n

(iii) Replacing x by 1 and a by x, we get (1 x)n = nC0

nC1x + nC2 x

2 ... + ( 1)n nCn x

n. (iv) Adding (1) and (2), we get (x + a)n +

(x a)n

= 2[xn + nC2 xn 2 a2 + nC4x

n 4 a4 + ...] = 2 (sum of terms at odd places). The last term is nCn a

n 1 or nCn 1 xan 1 since,

n is even or odd respectively. (v) Subtracting (2) from (1), we get, (x + a)n (x a)n = 2[nC1x

n 1 a + nC3xn

3 a3 + ...] = 2[sum of terms at even places] The last term is nCn 1 xa

n 1 or nCn an

according as n is even or odd respectively.

(vi) Number of terms in the expansion of (a) (x + a)n + (x a)n = n __ 2 + 1 when n is

even.

(b) (x + a)n (x a)n = n __ 2 when n is even

(c) (x + a)n (x a)n = ( n +1 ____ 2 ) when n is odd

(vii) Interchanging a and x in (1), we get, (a + x)n = nC0 a

n x0 + nC1 an 1 x1 + nC2 a

n

2 x2 + ... + nCr a

n r xr + ............. + nCn xn.

4. Important Results In the Binomial expansion of (x + a)n, if the sum of odd terms be A and the sum of even terms be B, then

(x + a)2n + (x a)2n = (A + B)2 + (A B)2 = 2(A2 + B2)

(x + a)2n (x a)2n = (A + B)2 (A B)2 = 4AB

(x2 a2)n = (A + B) (A B) = A2 B2

SOLVED SUBJECTIVE PROBLEMS: (CBSE/STATE BOARD): FOR BETTER UNDERSTANDING AND CONCEPT BUILDING OF THE TOPIC

1. If coefficient of x2 and x3 in the expansion of [3 + ax]9 are equal then find the value of a.

[NCERT] Solution Tr + 1 =

nCrxn rar Tr + 1 = 9Cr(3)9 r(ax)r

Tr + 1 = 9Cr(3)9 rarxr For coefficient of x2, r = 2 T2 + 1 = 9C239 2a2x2 T3 = (9C237a2)x2 Again, Tr + 1 =

9Cr(3)9 1(ax)r

For coefficient of x3 T3 + 1 = 9C3(3)

9 3a3x3

T4 = 9C3(3)6a3x3 According question, 9C23

7a2 = 9C336a3

9 8 _____ 2 3 = 9 8 7 ________ 3 2 1 a

a = 3 3 _____ 7 a = 9 __ 7 .

2. By using the binomial expansion, expand (1 + x + x2)3

Solution

(1 + x + x2)3 = [(1 + x) + x2]3

= 3C0.(1 + x)3 + 3C1.(1 + x)

2.(x2) + 3C2. (1 + x). (x2)2 + 3C3.(x

2)3

= (1 + x)3 + 3(1 + x)2.x2 + 3(1 + x).x4 + x6

= [1 + 3C1.x + 3C2x

2 + 3C3.x3] + 3x2

[1 + 2x + x2] + 3x4(1 + x) + x6 = (1 + 3x + 3x2 + x3) + (3x2 + 6x3

+ 3x4) + (3x4 + 3x5) + x6

= (x6 + 3x5 + 6x4 + 7x3 + 6x2 + 3x + 1). 3. Using the binomial theorem, find the value

of (102)6.

Solution (102)6 = (100 + 2)6

Binomial Theorem A.5

= 6C0 (100)6 + 6C1 (100)

5 2 + 6C2 (100)4 (2)2 + 6C3 (100)

3 (2)3 + 6C4 (100)

2 (2)4 + 6C5 100 (2)5

+ 6C6 (2)6

= (100)6 + 12 (100)5 + 60 (100)4 + 160 (100)3 + 240 (100)2 + 19200 + 64

= 100000000000 + 12000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64

= 1126162419264 4. If the coefficients of ar 1, ar, ar + 1 in the

binomial expansion (1 + a)n are in arithmetic progression, prove that n2 n(4r + 1) + 4r2 2 = 0.

[NCERT]

Solution

The general term in the expansion of (1 + a)n is given by tr + 1 =

nCrar.

Therefore coefficients of ar 1, ar and ar + 1 in the expansion are nCr 1,

nCr and nCr + 1, respectively.

Now, n Cr 1, n

Cr and n Cr + 1 are in A.P.

2n Cr = Cr 1 + nCr + 1 nCr 1 _____ nCr

+ nCr +1 _____ nCr

= 2

n ! _________________ (r 1) ! .(n r + 1)! (r!).(n r)!

__________ n ! +

n ! _________________ (r 1) ! .(n r + 1)! (r!).(n r)!

__________ n ! = 2

r _________ (n r + 1) + n r _____ n + 1 = 2

[Q (n r + 1)! = (n r + 1).(n r)! and (r!) = r. (r 1)!]

r(r + 1) + (n r)(n r + 1) = 2(r + 1)(n r + 1)

n2 n(4r + 1) + 4r2 2 = 0. 5. Which is larger (1.01)1000000 or 10,000?

[NCERT]

Solution Splitting 1.01 and using binomial theorem,

the first few terms are (1.01)1000000 =

(1 + 0.01)1000000 = 1000000C0 + 1000000C1(0.01)

+ other positive terms = 1 + 1000000 0.01 + other positive terms = 1 + 10000 + other positive terms > 10000 Hence, (1.01)1000000 >10000

6. Find the term independentof x in the

expansion of ( x2 + 2 + 1 __ x2 ) 8 .

Solution

( x2 + 2 + 1 __ x2 ) 8 . = [ ( x + 1 __ x ) 2 ] 8 = ( x + 1 __ x ) 16

Suppose, (r + 1)th term is independent of x

Now, Tr + 1 = nCr x

n r. ar = 16Cr.x16 r. ( 1 __ x )

r

= 16Cr. x16 2r

Q (r + 1)th term is independent of x x16 2r = x0

16 2r = 0 r = 8 Thus term independent of x is T9 = 16C8. 7. Find the fourth root of 624, correct to four

places of decimal.

Solution 6241/4 = (625 1)1/4 = (54 1)1/4

= 5 ( 1 1 __ 54 ) 1/4

= 5 [ 1 1 __ 4 1 __ 54 + 1 __ 4 ( 1 __ 4 1 ) ________ 2 ! ( 1 __ 54 ) 2 ] = 5 1 __ 4

1 __ 53 1 __ 4 .

3 __ 4 _____ 2 1 __ 57

= 5 2 ___ 103 12 ___ 127 = 5 .002 .0000012

= 4.9979988 = 4.9979

8. Expand ( x __ 3 + 1 __ x ) 5

[NCERT]

Solution Using Binomial theorem for positive

integral index, we have,

( x __ 3 + 1 __ x ) 5


= 5C0 ( x __ 3 ) 5 + 5C1 ( x __ 3 ) 4 ( 1 __ x ) + 5C2 ( x __ 3 ) 3 ( 1 __ x )

2 + 5C3 ( x __ 3 ) 2 ( 1 __ x )

3

+ 5C4 ( x __ 3 ) 1 ( 1 __ x ) 4 + 5C3 ( 1 __ x )

5

= ( x5 ____ 243 ) + 5 ( x4 ___ 81 ) ( 1 __ x ) + 10 ( x

3 ___ 27 ) ( 1 __ x2 )

+ 10 ( x __ 9 ) 2 ( 1 __ x3 ) + 5 ( x __ 3 ) ( 1 __ x4 ) + 1 __ x5 = x

5 ____ 243 +

5 ___ 81 x3 + 10 ___ 27 x +

10 ___ 9 ( 1 __ x ) + 5 __ 3 ( 1 __ x3 ) + 1 __ x5 9. Using Binomial theorem, indicate which

number is larger (1.1)10000 or 1000.[NCERT]

Solution Now (1.1)10000 = (1 + 0.1)10000 = 10000C0 + 10000C1(0.1) +

10000C2(0.1)2 + .... + 10000C10000(0.1)

10000

= 1 + 10000 1 ___ 10 + some positive terms

= 1 + 1000 + some positive terms > 1000 Hence, (1.1)10000 is larger than 1000.

10. Prove that r = 0

n

3r nCr = 4n. [NCERT]

Solution

Now r = 0

n

3r nCr r = 0

n

nCr 3r

= nC0 + vC1 3 + nC2 3

2 + ........ + nCn 3n = (1

+ 3)n = 4n

(Q (1 + x)n = nC0 + nC1x + nC2 x2 + ....... + nCn x

n) 11. Write the general term in the expansion of

(x2 yx)12, x 0.[NCERT]

Solution The given power of binomial is (x2 yx)12 is

{x2 + ( yx)}12

Here, the general term is Tr + 1 = 12Cr

(x2)12 r ( yx)}r = 12Cr x24 2r ( 1)r yr xr

= 12Cr ( 1)r x24 r yr.

12. Find a positive value of m for which the coeffi cient of x2 in the expansion (1 + x)m is 6.

[NCERT]

Solution Now (1 + x)m = mC0 +

mC1 + mC2 x

2 + ...... + mCmxm

we are given that coeffi cient of x2 = 6 mC2 = 6 m(m 1) ________ 2 ! = 6 m2 m = 6 2! = 6 2

= 12 m2 m 12 = 0 (m 4) (m + 3) = 0 m = 4, 3 But, m cannot be negative, therefore, m = 4. 13. Find a, b and n in the expansion of (a + b)n,

if the first three terms of its expansion are 729, 7290 and 30375, respectively.

[NCERT]

Solution It is given that T1 = 729 nC0 an b0 = 729 an = 729

(1) T2 = 7290 nC1 an 1 b1 = 7290 nan 1 b = 7290

(2) and T3 = 30375 nC2 an 2 b2 = 30375

n(n 1) _______ 2 an 2 b2 = 30375 (3) Multiplying (1) and (3), we get

n(n 1)

_______ 2 a2n 2 b2 = 729 30375 (4)

Squaring (2), we get n2 a2n 2 b2 = 7290 7290 Dividing (4) by (5), we have

n(n 1)

_______ 2n2 = 729 30375 ___________ 7290 7290

n 1 _____ 2n = 30375 ______ 72900 =

5 ___ 12

12n 12 = 10n 2n = 12 n = 6


Substituting, n = 6 in (1), we get a6 = 729 a6 = 36 a = 3 (6) Substituting, n = 6 and a = 3 in (ii), we

have

6(3)6 1 b = 7290 b = 7290 _____ 6 35 = 7290 _______ 6 243 b = 5.

Hence, the required power of binomial = (a + b)n = (3 + 5)6.

UNSOLVED SUBJECTIVE PROBLEMS: (CBSE/STATE BOARD): TO GRASP THE TOPIC, SOLVE THESE PROBLEMS

Exercise I

1. Expand (1 2x)5 by the binomial theorem. [NCERT] 2. Expand ( 2 __ x x __ 2 )

5 by the binomial theorem.

[NCERT] 3. Expand (2x 3)6 by the binomial theorem. [NCERT] 4. Expand ( x + 1 __ x ) 6 by the binomial theorem.

[NCERT] 5. Using binomial theorem, evaluate (96)3.

[NCERT] 6. Find a if the 17th and 18th terms of the

expansion of (2 + a)50 are equal.[NCERT]

7. The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1:7:42. Find n.

[NCERT] 8. Expand (x + y)5.

9. Expand the following (1 x + x2)4. 10. Expand the following expressions.

(i) (1 x)6

(ii) ( x 1 __ y ) 11 , y 0

Exercise II

1. Expand (x2 + 2y)5 by the binomial theorem.

2. Expand ( 2x 3 __ y ) 5 by the binomial theorem. 3. Find the value of r, if the coefficients of

(2r + 4)th and (r 2)th terms in the expansion of (1 + x)18 are equal.

4. Using binomial theorem, evaluate (101)4. [NCERT] 5. Using binomial theorem, evaluate (99)5. [NCERT] 6. Find the coeffcient of x6 y3 in the expansion

of (x + 2y)9.[NCERT]

7. Find the number of terms in the expansions of the following.

(2x 3y)9

8. Find the 7th term in the expansion

of ( 4x ___ 5 5 ___ 2x ) 9

9. If the coeffi cients of (r 1)th, rth and (r + 1)th terms in the expansion of (x +1)n are in the ratio 1 : 3 : 5 find n and r.

[NCERT] 10. Expand (x2 + 2a)5 by binomial theorem. 11. The 3rd, 4th and 5th terms in the expansion

of (x+a)n are, respectively 84, 280 and 560. Find the values of x, a and n.

ANSWERS

Exercise I

1. 1 10x + 40x2 80x3 + 80x4 32x5

2. 32 ___ x5 40 ___ x3 +

20 ___ x 5x + 5 __ 8 x

3 x5 ___ 32

3. 64x6 576x5 + 2160x4 4320x3 + 4860x2 2916x + 729.

4. x6 + 6x4 + 15x2 + 20 + 15 ___ x2 + 6 __ x4 +

1 __ x6


5. 884736 6. a = 1 7. n = 55 8. x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5

9. 1 4x + 10x2 16x3 + 19x4 16x5 + 10x6 4x7 + x8.

10. (i) 1 6x + 15x2 20x3 + 15x4 6x5 + x6

(ii) x11 11x10 y 1 + 55x9 y 2 165x8 y 3 + 330x7 y 4 462x6 y 5 + 462x5 y 6

330x4 y 7 + 165x3 y 8 55x2 y 9 + 11xy 10

Exercise II

1. x10 + 10x8y + 40x6y2 + 80x4y3 + 80x2y4 + 32y5

2. [ 32x5 240x4 _____ y + 720x3 _____ y2 1080x2 ______ y3 + 810x _____ y4 243 ____ y5 ] 3. r = 6 4. 104060401 5. 9509900499 6. 672 7. 10 terms

8. 10500 ______ x3 9. n = 7 and r = 3 10. x10 10x8a + 40x6a2 + 80x4a3 + 80x2a4 + 32a5

11. x = 1, a = 2 and n = 7

SOLVED OBJECTIVE PROBLEMS: HELPING HAND

1. Subjective the square root of 999 correct to three decimal places is

(a) 31.607 (b) 31.706 (c) 32.607 (d) 32.706

Solution

(a) 9991/2 = (900 + 99)1/2 = 9001/2 ( 1 + 99 ____ 900 ) 1/2

= 30 (1 + .11)1/2

= 30 [ 1 + 1 __ 2 (.11) + 1 __ 2 ( 1 __ 2 1 ) ________ 2 ! (.11)2 +

1 __ 2 ( 1 __ 2 1 ) ( 1 __ 2 2 ) _______________ 3 ! (.11)2 + ] = 30 [ 1 + . 11 ____ 2 1 __ 8 (. 11)2 + 1 __ 6 (. 11)3 ] = 30 [1 + 0.055000 0.001512 + 0.000083....] = 30 [1.053571] = 31.60713 = 31.607 (Correct to three places of

decimal) 2. The positive integer just greater than

(1 + 0.0001)10000 is (a) 4 (b) 5 (c) 2 (d) 3

[AIEEE 2002]

Solution (d) We know that e = li m n ( 1 + 1 __ n )

n and

2 < e < 3 (1 + 0.0001)10000 < 3 (By putting n = 10000) Also, (1 + 0.0001)10000 = 1 + 10000 10 4

+ 1000 9999 ___________ 2 ! 108 + .... upto 10001 terms

(1 + 0.0001)10000 > 2. Hence, 3 is the positive integer just greater

than (1 + 0.0001)10000 > 2. Hence, (d) is the correct option. 3. If the coefficients of second, third and

fourth term in the expansion of (1 + x)2n are in A.P., then 2n2 9n + 7 is equal to

(a) 1 (b) 0 (c) 1 (d) 3/2[AMU 2001; MP PET 2004]

Solution T2 =

2nC1 x, T3 = 2nC2 x

2, T4 = 2nC3 x

3

Coeffi cient of T2, T3, T4 are in A.P.

2. 2nC2 = 2nC1 + 2nC3 2n ! __________ 2 !(2n 2) !

= 2n ! ________ (2n 1) ! + 2n ! __________ 3 !(2n 3) !

2.2n(2n 1) __________ 2 = 2n + 2n(2n 1)(2n 1)

_______________ 6


n(2n 1) = n + (n) (2n 1)(2n 2) ________________ 6 6(2n2 n) = 6n + 4n3 6n2 + 2n 6n(2n 1) = 2n(2n2 3n + 4) 6n 3 = 2n2 3n + 4 0 = 2n2 9n + 7 2n2 9n + 7 = 0. 4. Subjective the value of is equal to (a) 1 (b) 2 (c) 2 (d) 4

(183 + 73 + 3.18.7.25)

___________________________ 36 + 6.243.2 + 15.81.4 + 20.27.8 .

+ 15.9.16 + 6.3.32 + 64

Solution (a) The numerator is of the form a3 + b3 + 3ab (a + b) = (a + b)3 where a = 18 and b = 7.

Therefore, Nr = (18 + 7)3 = 253

For Dr, 31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243

6C1 = 6, 6C2 = 15,

6C3 = 20, 6C4 =

6C2 = 15, 6C5 =

6C1 = 6, 6C6 = 1

Dr = 36 + 6C135 . 21 + 6C234 . 22 + 6C333. 23 + 6C5 3.2

5 + 6C626.

This is clearly the expansion of (3 + 2)6 = 56 = (25)3

Nr ___ Dr = (25)3

_____ (25)3 = 1

5. The value of x for which the sixth term in the expansion of

[ 2log2 _____ 9x1 + 7 + 1 _______________ 21/5 log2(3x1 + 1) ] 7 is 84

is equal to (a) 1 (b) 2 (c) 1, 2 (d) 3

[DCE 1993, 1995]

Solution

(c) Exp. = [ _______ 9x 1 + 7 1 __________ (3x 1 + 1)1/5 ] 7 Now, T6 = 84

7C5 ( _______

9x 1 + 7 ) 2 ( 1 _________ (3x 1 +1)1/5 ) 5 = 84 21 (9x 1 + 7). 1 _________ (3x 1+ 1) = 84

9 x 1 + 7 = 4 (3x 1 + 1)

32x ___ 9 + 7 = 4 __ 3 3

x + 4 32x 12.3x + 27 = 0 (3x 3) (3x 9) = 0 3x = 3 or 3x = 9 x = 1, 2 6. If the coefficient of mth, (m + 1)th and

(m + 2)th terms in the expansion of (1 + x)m are in AP, then

(a) n2 + n(4m + 1) + 4m2 2 = 0 (b) n2 + n(4m + 1) + 4m2 + 2 = 0 (c) (n 2m)2 = n + 2 (d) (n + 2m)2 = n + 2 [AIEEE 2005] Solution (c) 2.nCm =

nCm 1 + nCm + 1

2. n ! ___________ m ! (n m) ! = n ! __________________ (m 1) ! (n m + 1) !

+ n ! __________________ (m + 1) ! (n m + 1) !

2(m + 1) (n m + 1) = (m + 1)m + (n m + 1) (n m)

4mn 4m2 + n n2 + 2 = 0 (n 2m)2 = n + 2 7. 1 1! + 2 2! + 3 3! + ... + n n! is equal to (a) (n + 1)! (b) (n + 1)! 1 (c) (n + 1) ! + 1 (d) none of these

Solution (b) 1 1! + 2 2! + 3 3! + .... + n n !

= r = 1

n

rr ! = r = 1

n

{(r + 1) 1} r !

= r = 1

n

{(r + 1)r ! r !}

= r = 1

n

((r + 1) ! r !)

= (2! 1!) + (3! 2!) + (4! 3!) + ... + ((n + 1)! n!)

= (n + 1) ! 1! = (n + 1)! 1 8. The value of {1.3.5.....(2n 3) (2n 1)} is (a) (2n)!/n! (b) (2n)!/2n (c) n!/(2n)! (d) None of these

Solution (a) 1.3.5........(2n 1)2n =


1.2.3.4.5.6.....(2n 1)(2n)2n

_______________________ 2.4.6.....2n

= (2n) ! 2n

___________ 2n(1.2.3.n) = (2n) !/n !

9. If in the binomial expansion of [

_______ 2log(103x) + 5

_______ 2(x 2)log3 ] m 6th term is equal

to 21 and coefficients of 2nd, 3rd, 4th terms are 1st, 3rd and 5th term of A.P., then the value of x (where log is defined at the base of 10)

(a) 0 (b) 1 (c) 2 (d) 3[MPPET 2007]

Solution (c) As the coefficients mC1,

mC2 and mC3 of

T2, T3, T4 are the first, third and fifth term of an A.P. whose common difference is 2d, therefore 2 . mC2 =

mC1 + mC3 (m 2)

(m 7) = 0 As the sixth term is 21 and m = 2 violates

the rule, therefore, we will take m = 7 and T6 = 21

= 7C5 [ _______

2log(103x) ] 7 5 [ 5 _______

2(x 2)log 3 ] 5 = 21 = 21.2log(10 3x) + log 3x 2 = 2log[(10 3x)3x 2]

= 1 = 20

On solving, we get, x = 0, 2. 10. In the expansion of (2 3x3)20, if the ratio of

10th term of 11th term is 45/22, then x = (a) 2/3 (b) 3/2 (c) 2/3 (d) 3/2

[Orissa JEE 2007] Solution (c) Given expansion (2 3x3)20

Tr + 1 = 20Cr 220 r ( 3x3)r putting r = 9, 10 t10 = 20C9 211 ( 3x3)9 t11 =

20C10 210 ( 3x3)10 Q

t10 __ t11 = 45 ___ 22

10 ___ 11 ( 2 __ 1 ) 1 ____ 3x3 = 45 ___ 22 ( Q 20 C9 ____ 10 C10 = 10 ___ 11 ) x3 = 8 ___ 27 x = 2 ___ 3 11. (

__ 3 + 1)4 + (

__ 3 1)4 is equal to

(a) a rational number (b) an irrational number

(c) a negative integer (d) none of these

Solution (a) (

__ 3 + 1)4 + (

__ 3 1)4

= 2 {4C0 ( __

3 )4 + 4C2 ( __

3 )2 + 4C4}, which is positive integer and hence a

rational number. 12. The coefficient of x53 in the following

expansion

m=0

100

100Cm(x 3)100 m.2m is

(a) 100C47 (b) 100C53

(c) 100C53 (d) 100C100[IIT Sc. 1992]

Solution (c) The given sigma is expansion of [(x 3) + 2]100 = (x 1)100 = (1 x)100

x53 will occur in T54. T54 = 100C53( x)53 Coefficient is 100C53. 13. If in the expansion of (1 + x)m(1 x)n,

the coefficient of x and x2 are 3 and 6 respectively, then m is

(a) 6 (b) 9 (c) 12 (d) 24[IIT 1999; MP PET 2000]

Solution (c) (1 + x) m (1 x)n

= [ 1 + mx + m (m 1) 2 __________ 2 ! + ] ( 1 nx + n(n 1) ___________ 2 ! x2 ) = 1 + (m n)x

+ ( n2 n _____ 2 mn + (m2 m)

_______ 2 ) x2 + Given m n = 3 or n = m 3

Hence, n2 n _____ 2 mn +

m2 m ______ 2 = 6

(m 3)(m 4) ____________ 2 m (m 3) + m2 m ______ 2

= 6 m2 7m + 12 2m2 + 6m + m2 m + 12 = 0 2m + 24 = 0 m = 12


14. If a1, a2, a3 are coeffi cients of any four consecutive terms in the expansion of

(b + x)n, then a1 ______ a1 + a2

+ a3 ______ a3 + a4

is equal to

(a) a2 ______ a2 + a3

(b) 2a2 ______ a2 + a3

(c) a2 ______ a2 + a3

(d) a2 _________ 2 (a2 + a3)

[IIT 1975]

Solution (b) Let given coefficients are those of

Tr, Tr + 1, Tr + 2, Tr + 3. Then

a2 __ a1 =

coef. of Tr+1 ___________ coef. of Tr =

Cr ____ Cr 1 = n r + 1 ________ r

1 + a2 __ a1 = n + 1 _____ r

a1 ______ a1 + a2 = r _____ n + 1

(1) Similarly,

a2 ______ a2 + a3 = r + 1 _____ n + 1 (2)

a3 ______ a3 + a4

= r + 2 _____ n + 1 (3)

Now, (1) + (3)

a2 ______ a1 + a2 + a3 ______ a3 + a4

= 2(r + 1)

_______ n + 1 = 2a2 ______ a2 + a3

[by (2)]

1. If the coefficient of 7th and 13th term in the expansion of (1 + x)n are equal, then n =

(a) 10 (b) 15 (c) 18 (d) 20 2. In the expansion of (1 x)5, coefficient of

x5 will be (a) 1 (b) 1 (c) 5 (d) 5 3. If the ratio of the coefficient of third and

fourth term in the expansion of ( x 1 ___ 2x ) n is

1: 2, then the value of n will be (a) 1 8 (b) 1 6 (c) 12 (d) 10 4. If the coefficients of rth term and (r + 4)th

term are equal in the expansion of (1 + x)20, then the value of r will be

(a) 7 (b) 8 (c) 9 (d) 10[MPPET 2002]

5. Sum of odd terms is A and sum of even terms is B in the expansion (x + a)n, then

[RPET 1987, 1992; UPSEAT 2004;Roorkee 1986]

(a) AB = 1 __ 4 [(x a)2n (x + a)2n]

(b) 2AB = (x + a)2n (x a)2n

(c) 4AB = (x + a)2n (x a)2n

(d) none of these 6. 9th term in the expansion of ( y __ 2 + 2x )

12 is

(a) 7920 x7x5 (b) 7920 x6y6

(c) 7920 x8y4 (d) 7816 x8x4

7. If A and B are the coefficient of xn in the expansions of (1 + x)2n and (1 + x) 2n 1 respectively, then

(a) A = B (b) A = 2B (c) 2A = B (d) none of these [NCERT] 8. The total number of terms in the expansion

of (x + a)100 + (x a)100 after simplification will be

(a) 202 (b) 51 (c) 50 (d) none of these 9. If p and q be positive, then the coeffi cient

of xp and xq in the expansion of (1 + x) p + q will be

(a) equal (b) equal in magnitude but opposite in sign (c) reciprocal to each other (d) none of these

[NCERT]

OBJECTIVE PROBLEMS: IMPORTANT QUESTIONS WITH SOLUTIONS


10. If the coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)n be in A.P.m then n =

(a) 7 only (b) 14 only (c) 7 or 14 (d) none of these

[Roorkee 1984] 11. The value of (

__ 5 + 1)5 (

__ 5 1)5

(a) 252 (b) 352 (c) 452 (d) 532

[MPPET 1985] 12. If the three consecutive coefficient in the

expansion of (1 + x)n are 28, 56 and 70, then the value of n is

(a) 6 (b) 4 (c) 8 (d) 10[MPPET 1985]

13. In the expansion of (x2 2x)10, the coeffi cient of x16 is

(a) 1680 (b) 1680 (c) 3360 (d) 6720 14. If T2/T3 in the expansion of (a + b)n, and T3/

T4 in the expansion of (a + b) n + 3 are equal,

then n = (a) 3 (b) 4 (c) 5 (d) 6 [RPET 1987, 1996]

15. If the coefficients of x7 and x8 in ( 2 + x __ 3 ) n

are equal, then n is

(a) 56 (b) 55 (c) 45 (d) 15 16. If the coefficient of (2r + 4)th and (r 2)th

terms in the expansion of (1 + x)18 are equal, then r =

(a) 12 (b) 10 (c) 8 (d) 6[PCET 2008; MPPET 1997]

17. If the second,third and fourth term in the expansion of (x + a)n are 240, 720 and 1080, resapectively, then the value of n is:

(a) 15 (b) 20 (c) 10 (d) 5[Kurukshetra CEE 1991; DCE 1995, 2001] 18. If the coefficients of Tr, Tr + 1, Tr + 2 terms

of (1 + x)14 are in A.P., then r = (a) 6 (b) 7 (c) 8 (d) 9

19. The expansion [ x + (x3 1 ) 1 __ 2 ] 5 + [ x + (x3 1 ) 1 __ 2 ] 5 is a polynomial of degree

(a) 5 (b) 6 (c) 7 (d) 8

[IIT 1992; DCE 1996, 2006] 20. In the expansion of (x + a)n, the sum of odd

terms is P and Sum of even terms is Q, then the value of (P2 Q2) will be:

(a) (x2 + a2)n (b) (x2 a2)n

(c) (x a)2n (d) (x + a)2n

[RPET 1997; Pb CET 1998]

SOLUTIONS

1. (c) nC6 = nC12 n = 6 + 12 n = 18 2. (b) (1 x)5, coeffi cient of x5

Tr + 1 = nCr an r br in the expansion of (1 x)5 We have Tr + 1 =

5Cr (1) 5 r ( x)r

= ( 1)r 5Cr xr

Q T r + 1 contains x5 r = 5 Hence, the coefficient of x5 in the expansion

= 5C5 ( 1)5

= 1 1 = 1 3. (d) T3 = nC2 (x)n 2 ( 1 ___ 2x )

2 and

T4 = nC3(x)

n 3 ( 1 ___ 2x ) 2

But according to the condition,

n(n 1) 3 2 1 8

______________________ n(n 1)(n 2) 2 1 4 = 1 __ 2 n = 10

4. (c) Tr + 1 = nCr xr for (1 + x)n

Here the coefficient is nCr.


Given nCr 1 = nCr + 3 20Cr 1 = 20Cr + 3

(r 1) + (r + 3) = 20 r = 9. 5. (c) (x + a)n = nC0xn + nC1xn 1a + nC2 xn 2a2

+ nC3xn 3a3 + ..... + an (1)

(x a)n = nC0xn nC1x

n 1a + nC2xn 2a2

nC3xn 3a3 + ..... + ( 1)nan

(2) By assumption, A = nC0x

n + nC2xn 2a2 + nC4x

n 4a4 + ........

B = nC1xn 1a + nC3x

n 3a3 + nC5xn 5a5 + ........

This A + B = (x + a)n, A B = (x a)n 4AB = (A + B)2 (A B)2 = (x + a)2n

(x a)2n. 6. (c) We know that in the expansion of

(a + b)n, we have (r + 1)th term Tr + 1 = nCr

an rbr

in the expansion of ( y __ 2 + 2x ) 12

We have 9th term, T9 = T8 + 1

= 12C8 ( y __ 2 ) 12 8

(2x)8 [Here a = y __ 2 , b = 2x]

= 12C8 ( y __ 2 ) 4 (2x)8

= 12C8 y4

__ 24 28 x8

= 12C4 24 x8 y4

= 12 11 10 9 16 __________________ 4 3 2 1 x8y4

= 72 100 x8y4

= 7920 x8y4 7. (b) A = Coefficient of xn in (1 + x)2n = 2nCn. B = coefficient of xn in (1 + x)2n 1 = 2n 1 Cn =

2n 1C(2n 1) n = 2n 1Cn 1

Now A = ( 2n ___ n ) (2n 1Cn 1) = 2B 8. (b) (x + y)100 + (x y)100 = 2[C0xn + C2xn

2 y2 + C4xn 4 y4 + .......... + Cny

n], where n = 100

Total terms = ( 100 ____ 2 ) + 1 =51. 9. (a) Tr + 1 = p + qCr xr

coefficient of xr = p + qCr.

Hence, coefficient of xp = p + qCp and that of xq is p + qCq.

Note that p + qCp = p + qCq as

nCr = nCn r.

10. (c) Coefficient of T5, T6, T7 are in A.P. for(1 + x)n

nC4, nC5, nC6 are in A.P. 2(nC5) = nC4 + nC6 2n ! __________ 5 ! (n 5) ! =

n ! __________ 4 ! (n 4) ! = n ! __________ 6 ! (n 6) !

2 ________ 5 (n 5) = 1 ____________ (n 4) (n 5) +

1 ___ 6.5

If we put n = 7 in (1), 2 _____ 5 (2) = 1 ___ 3.2 +

1 ___ 30 or 1 __ 5

= 1 __ 5 (True) n = 14 in (1) 2 _____ 5 9 = 1 ______ 10 9 + 1 ___ 30 (True) 11. (b) (

__ 5 + 1)5 (

__ 5 1)5 = (1 +

__ 5 )5

+ (1 __

5 )5

= 2[1 + 5C2( __

5 )2 + 5C4( __

5 )4]

= 2 [ 1 + 5.4 ___ 2 ! .5 + 5 (52) ] = 352 12. (c) Coefficient in Tr + 1 = 28, Tr + 2 = 56, Tr + 3

= 70. This nCr = 28 (1) nCr + 1 = 56 (2) nCr + 3 = 70 (3) Here we apply

nCr _____ nCr 1

= n (r 1)

_________ r ,

we get

( 2 )

____ ( 1 ) nCr + 1 _____ nCr

= 56 ___ 28 n r _____ r + 2 = 2 n = 3r + 2 (4)

( 3 )

____ ( 2 ) nCr + 2 _____ nCr + 1

= 70 ___ 56 n (r + 1)

_________ r + 2 = 5 __ 4

(3r + 2) (r + 1) ______________ r + 2 = 5 __ 4 r = 2 n = 8 by (4) 13. (c) (x2 2x)10 = x10 (x 2)10 (1)

For coefficient of x16 in (1), we consider coefficient of x6 in

(x 2)10, Tr + 1 = 10Cr (x)

10 r ( 2)r (2) 10 r = 6, r = 4 By (2), coefficient = 10C4( 2)

4 = 10.9.8.7 _______ 4 ! .(16)

= 3360

14. (c) For (a + b)n: T2 __ T3

= nC1a

n 1b ________ nC2a

n 2b2 = ( 2 _____ n 1 ) ( a __ 2 ) (1)

For (a + b)n + 3: T3 __ T4

= n + 3C2a

n + 3 2b2 ___________ n + 3C3a

n + 3 2b3

= (n + 3) (n + 2) .3 !

______________________ 2 ! . (n + 3) (n + 2) (n + 1) . ( a __ b )

or, T3 __ T4

= ( a __ b ) ( 3 _____ n + 1 ) (2) By asssumption and by (1) and (2),

( a __ b ) ( 2 _____ n 1 ) = ( a __ b ) ( 3 _____ n + 1 ) or, 2 _____ n 1 = 3 _____ n + 1

or, 2n + 2 = 3n 3 or, n = 5.

15. (b) Tr + 1 = nCr 2 n r ( x __ 3 ) r = nCr

2n r ____ 3r xr.

We are given nC7 2n 7 ____ 37 =

nC8 2n 8 ____ 38

nC8 ___ nC7

= 6 n ! __________ (n 8) ! 8 ! . (n 7) ! 7 !

__________ n ! = 6

n 7 _____ 8 = 6 n =55. 16. (d) Coefficient in T2r + 4 = Coeffi cient in

T(r 2) for (1 + x)18

This 18C2r + 3 = 18Cr 3 (2r + 3) + (r 3) = 18 r = 6 17. (d) T2 = nc1xn 1a = 240, T3 =

nc2xn 2a2 = 720 T4 =

nc3xn 3a3 = 1080

nxn 1a ____________ n (n 1)

_______ 2 xn 2a2

= 240 ____ 720 2x _______ (n 1)a = 1 __ 3

(1)

and n (n 1)

________ 1.2 xn 2a2 ___________________

n (n 1) (n 2)

_____________ 1.2.3 xn 3a3

= 720 _____ 1080

x _______ (n 2)a = 2 __ 3 (2)

Divide (1) by (2) 2 (n 2)

________ 3(n 1) = 1 __ 2

4n 8 = 3n 3 n = 5 18. (d) Tr = 14cr 1xr 1; Tr + 1 = 14crxr; Tr + 2 = 14cr +

1xr + 1

By the given condition

2 . 14cr = 14cr 1 +

14cr + 1

2. 14 ! __________ r ! (14 r) ! = 14 ! _______________ (r 1) ! (15 r) !

+ 14 ! _______________ (r + 1) ! (13 r) !

2 _______________________ r (r 1) ! (14 r) (13 r) !

= 1 ____________________________ (r 1) ! (15 r) (14 r) (13 r) !

2 ________ r (14 r) = 1 _____________ (15 r) (14 r) +

1 ______ (r + 1)r

+ 1 ______________________ (r + 1) r (r 1) ! (13 r) !

(15 r) r ______________ r (5 r) (14 r) = (14 r) (r + 1)

______________ (r + 1) r (14 r)

15 2r ______ 15 r = 13 2r ______ r + 1

15r + 15 2r2 2r = 195 30r 13r + 2r2 4r2 56r + 180 = 0 r2 14 r + 45 = 0 (r 5) (r 9) = 0 r = 5, 9 But 5 is not given. Hence, r = 9. 19. (c) [x + (x3 1)1/2]5 + [x (x3 1)1/2]5

= 2 [5C0x5 + 5C2 x

3 (x3 1) + 5C4 x1

(x3 1)2] Max. power of x is 7. 20. (b) C0 xn + C2 xn 2 a2 + C4 xn 4 a4 + ...... = 0

= Sum of odd terms = p C1x

n 1a + C3xn 3a3 + C5x

n 5a5 + ..... = E = sum of even terms = q

This p + q = (x + a)n, p q = (x a)n p2 q2 = [(x + a) (x a)]n = (x2 a2)n.



UNSOLVED OBJECTIVE PROBLEMS (IDENTICAL PROBLEMS FOR PRACTICE): FOR IMPROVING SPEED WITH ACCURACY

1. If the coefficients of (2r + 1)th and (r + 5)th terms in the expansion of (1 + x)25 are equal, then the value of r is:

(a) 4 or 7 (b) 4 or 6 (c) 4 (d) 6 2. If in the expansion of (1 x)n the coeffi cient

of x2 be 3, then the values of n are: (a) 3, 2 (b) 3, 2 (c) 3, 2 (d) 3, 2 3. If for positive integers r > 1, n > 2, the

coefficient of the (3r)th and (r + 2)th powers of x in the expansion of (1 + x)2n are equal, then

(a) n = 2r (b) n = 3r (c) n = 2r + 1 (d) none of these 4. The number of non-zero terms in the

expansion of (1 + 3 __

2 x)9 + (1 3 __

2 x)9 is

(a) 9 (b) 0 (c) 5 (d) 10[EAMCET 1991]

5. If coefficient of (2r + 3)th and (r 1)th terms in the expansion of (1 + x)15 are equal, then value of r is

(a) 5 (b) 6 (c) 4 (d) 3[RPET 1995, 2003; UPSEAT 2001]

6. If coefficients of 2nd, 3rd and 4th terms in the binomial expansion of (1 + x)n are in A.P., then n2 9n is equal to

(a) 7 (b) 7 (c) 14 (d) 14[RPET 1999; UPSEAT 2002]

7. The coefficients of three successive terms in the expansion of (1 + x)n are 165, 330 and 462 respectively, then the value of n will be

(a) 11 (b) 10 (c) 12 (d) 8 8. If the coefficient of 4th term in the

expansion of (a + b)n is 56, then n is (a) 12 (b) 10 (c) 8 (d) 6

[AMU 2000] 9. The coefficient of x5 in the expansion of

(x + 3)6 is (a) 18 (b) 6 (c) 12 (d) 10

[DCE 2002] 10. In the expansion of (1 + x)n, coefficients

of 2nd, 3rd, and 4th terms are in A.P., then n is equal to

(a) 7 (b) 9 (c) 11 (d) none of these 11. What is the approximate value of (1.02)8 ? (a) 1.171 (b) 1.175 (c) 1.177 (d) 1.179 [NDA 2008] 12. The coefficient of x12 in the expansion of

(x2 + 2x)10 is: (a) 11520 (b) 13410 (c) 16520 (d) 23040

[SCRA 2007]

WORK SHEET: TO CHECK PREPARATION LEVEL

Important Instructions: 1. The answer sheet is immediately below the

work sheet. 2. The test is of 8 minutes. 3. The test consists of 8 questions. The maximum marks are 24. 4. Use blue/black ball point pen only for

writing particulars/marking responses. Use of pencil is strictly prohibited.

1. If coefficients of (2r + 1)th term and (r + 2)th expansion of (1 + x)43, then the value of r will be

(a) 14 (b) 15 (c) 13 (d) 16

[UPSEAT 1999]


2. In the expansion of (1 + x)11, the fifth the third term. Then the value of x2 is

(a) 4 (b) 9 (c) 16 (d) 24 3. After simplifi cation, what is the number

of terms in the expansion of [(3x + y)5]4 [(3x + y)4]5 ?

(a) 4 (b) 5 (c) 10 (d) 11

[NDA 2007] 4. In the expansion of (1 + x)n the coefficient

of pth and (p + 1)th terms are respectively p and q. Then p + q =

(a) n + 3 (b) n + 1 (c) n + 2 (d) n 5. The first 3 terms in the expansion of (1 + ax)n

(n 0) are 1, 6x and 16x2. Then, the value of a and n are; respectively

(a) 2 and 9 (b) 3 and 2

(c) 2/3 and 9 (d) 3/2 and 6 [Kerala Engg. 2002] 6. If tr is the rth term in the expansion of

(1 + x)101, then what is the ratio t20 __ t19

equal to?

(a) 20x ____ 19 (b) 83x

(c) 19x (d) 83x ____ 19

[NDA 2008]

7. What is the coefficient of x3y4 in (2x + 3y2)5 ? (a) 240 (b) 360 (c) 720 (d) 1080 [NDA 2008] 8. The coefficient of the (m + 1)th term and

the (m + 3)th in the expansion of (1 + x)20 are equal then value of m is

(a) 10 (b) 8 (c) 9 (d) none of these [UP-SEE 2007]

ANSWER SHEET

1. a b c d 4. a b c d 7. a b c d 2. a b c d 5. a b c d 8. a b c d 3. a b c d 6. a b c d

HINTS AND EXPLANATIONS

an = 6 and n (n 1) ________ 2 a2 = 16 Solving, a = 2 __ 3 and n = 9

6. b20 ___ b19

= 101C19x

19

_______ 101C18x18 =

101 19 + 1 ___________ 19 = 83x ____ 19

1. T2r + 1 = Tr + 2 43C2r = 43Cr + 1 2r = r + 1 or 2r + r + 1 = 43 r = 1 or r = 14 3. ((3x + y)5)4 [(3x y)4)5 = (3x + y)20

(3x y)20

Number of terms = 20 ___ 2 = 10 5. T1 = 1, T2 = 6x, T3 = 16x2 nC0 (ax)

0 = 1; nC1 (ax)1 = 6x; nC2 (ax)

2 = 16x2

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