Loading....
Coupon Accepted Successfully!

 

Question-1

Expand each of the following expression (1 – 2x)5

Solution:
By using Binomial Theorem, we have
(1 –2x)5 = [1+(-2x)]
5
                  = 5C0 + 5C1(-2x) + 5C2(-2x)2 + 5C3(-2x)3 + 5C4(-2x)4 + 5C5(-2x)5
                  = 1 + 5 (-2x) + 10 (-2x )2 + 10 (-2x)3 + 5(-2x)4 + (-2x)5
                  = 1 – 10x + 40x2 - 80x3 + 80x4 - 32x
5

Question-2

 Expand each of the following expression

Solution:
By using Binomial Theorem, we have

=

          = 5C05C1 + 5C2+ 5C3 + 5C4+ 5C5

          =

          =

Question-3

Expand each of the following expression (2x – 3)6

Solution:
By using Binomial Theorem, we have
(2x -3)6= [2x +(-3)]
6
           = 6C0 (2x)6 (-3)° + 6C1(2x)5 (-3)1 + 6C2 (2x)4 (-3)2 + 6C3(2x)3 (-3)3+ 6C4 (2x)2 (-3)4 + 6C5(2x)1 (-3)5 + 6C6(2x)0 (-3)6
           = 1(2x)6 + 6(2x)5(-3) + 15(2x)4 (-3)2 + 20(2x)3 (-3)3 + 15(2x)2(-3)4+ 6(2x)(-3)5+(-3)6
           = 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

Question-4

Expand each of the following expression

Solution:
By using Binomial Theorem, we have;
= 5C0 + 5C1 + 5C2 + 5C3 + 5C4 +5C5

           =

           =

Question-5

Expand each of the following expression

Solution:
By using Binomial Theorem, we have:

= 5C0(x)6+6C1(x)5+ 6C2(x)4+ 6C3(x)3+ 6C4(x)2+
6C5(x)1+6C6

             = x6 + 6x5

          
l = x6 + 6x4 + 15x2 + 20 +

Question-6

Using binomial theorem, evaluate each of the following (96)3

Solution:
We express 96 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem
Write 96 = 100 - 4
Therefore
(96)3 = (100 - 4)
3
        = 3C0 (100)3 - 3C1(100)2(4) + 3C2(100)1(4)2 - 3C3(4)3
        = 1000000 -3 (10000) (4) + 3 (100) (16) - (64)
        = 1000000 - 120000 + 4800 - 64
        = 884736

Question-7

Using binomial theorem, evaluate each of the following (102)5

Solution:
(102)5 = (100 + 2)5

          = 5C0(100)5 + 5C1(100)4(2) + 5C2(100)3(2)2 + 5C3(100)2(2)3 + 5C4(100)(2)4 + 5C5(2)5

          =10000000000 + 5(100000000)(2)+10(1000000)(4)+ 10(10000)(8) + 5(100)(16) + 32

          = 10000000000+1000000000 + 40000000 + 800000 + 8000 + 32

          = 11040808032

Question-8

Using binomial theorem, evaluate each of the following (101)4

Solution:
(101)4 = (100+ 1)4

          = 4C0(100)4 + 4C1(100)3(1) + 4C2(100)2(1)2 +4C3(100)1(1)3 +4C4(1)4

          = 100000000 + 4(1000000) + 6 (10000)+ 4 (100)+ 1

          = 100000000 + 4000000 + 60000 + 400+1

          = 104060401

Question-9

Using binomial theorem, evaluate each of the following (99)5

Solution:
(99)5 = (100- 1)5
            = 5C0(100)5 - 5C1(100)4.1 + 5C2(100)3.(1)2 - 5C3(100)2.(1)2 + 5C4(100)1(1)4 - 5C5(1)5
            = (100)5 – 5
× (100)4 + 10 ×(100)3 – 10 × (100)2 + 5 × 100 – 1
            = 10000000000-500000000+ 10000000 - 100000 + 500 - 1
            = 10010000500 - 500100001
            = 9509900499

Question-10

Using Binomial Theorem indicate which number is larger (1.1)10000 or 1000.

Solution:
Splitting 1.1 and using Binomial Theorem to write the first few terms we have

(1.1)10000 = (1+0.1)10000
              = 10000C0 + 10000C1(0.1) +10000C2(0.1)2 + other positive terms.
              = 1 + 10000 ´ (0.1) + other positive terms
              = 1 + 1000 + other positive terms
              >1000
Hence, (1.1)10000 > 1000.

Question-11

Find (a + b)4 - (a – b)4. Hence, evaluate .

Solution:
(a + b)4 -(a - b)4
= [4C0a4 + 4C1a3b + 4C2a2b2 + 4C3ab3 + 4C4b4] - [4C0a4 + 4Cla3b + 4C2a2b2 - 4C3 ab3 + 4C4 b4] = 2 × 4C1 a3b + 2 × 4C3 ab3
= 2[4a3b + 4ab3]
= 8ab [a2 + b2]
Thus, ( +)4 - ( - )4
= 8 . [3 + 2] = 8(5) = 40

Question-12

Find (x + 1)6 + (x - 1)6. Hence or otherwise evaluate

Solution:
(x + 1)6 + (x - 1)6
= [6C0 x6 + 6C1 x5 + 6C2 x4] + [6C3 x3 + 6C4 x2 + 6C5 x1 - 6C6] +
[6C0 x6 - 6C1 x5 + 6C2 x4 - 6C3 x3 + 6C4 x2 + 6C5x1 6C6]
= 2[6C0x6 + 6C2 x4 + 6C4 x2 + 6C6]
= 2[x6+ 15x4+ 15x2+ 1]
Thus, (+ 1)6 + (- 1)6
= 2 [()6 + 15()4 + 15 () + 1]
= 2 [8+ 15 (4)+ 15 (2) + 1]
= 2 [8 + 60 + 30+ 1]= 198.

Question-13

Show that 9n+1 - 8n - 9 is divisible by 64, whenever n is a positive integer.

Solution:
n = 1 9n + 1 - 8n - 9 = 92 - 8 - 9
                                = 81 - 17 = 64= 1(64)
n = 2
9n + 1 - 8n - 9 = 93 - 8(2) - 9
                                = 729 – 16 - 9 = 704= 11 (64)
From n = 3, 4, 5,.....9n + 1 – 8n - 9 = 9(1 + 8)n - 8n - 9
                                                 = 9 [nC0 + nC1 . 8 + nC2.82 + ... nCn 8n] – 8n - 9
                                                 = 9[1 + 8n + nC2.82 + ... nCn 8n] –8n – 9
                                                 = 9 + 72n + 9. nC2. 82 + ... 9 nCn 8n –8n - 9
                                                 = 82 [n + 9 (nC2 + nC3.8 +... nCn 8n-2)]
                                                           which is divisible by 64.

Question-14

Prove that 3r nCr = 4n .

Solution:
L.H.S = 30C(n,0) + 31 C(n,1) + 32 C(n,2) + --- 3r(n, r) + --- +3n C(n, n) 
        = C(n,0) + C(n,1) 31 + C(n,2) 32 + C(n,3) 33 + --- +C(n,n)3n
This is in the form of (1+3)n  
        = (1+3)n = 4n = R.H.S

Question-15

Prove that x5 in (x + 3)8

Solution:
Suppose x5 occurs in the (r + 1 )th term of the expansion (x + 3)8
Now Tr+1 = nCr an - r br = 8Cr x8 - r 3r
Comparing the indices of x in x5 and in Tr + 1, we get r = 3
Thus, the coefficient of x5 is

8C3(3)3 = 1512

Question-16

Prove that a b7 in (a-2b)12

Solution:
Let a5b7 occurs in the (r + 1)th term, in the expansion of (a - 2b)12 given by I2Cr. a12 – r(-2b)r. Then 12 – r = 5. This gives r= 7. 
Thus the coefficient of a5b7 is


12C5 (-2)7 = (-128) = (792) (-128) = -101376

Question-17

Prove that (x2 – y)6

Solution:
We have Tr + 1 in (a + b)n = nCran - r. br ,0 r n
Tr + 1 in (x2 - y)6 = 6Cr(x2)6 - r (-y)r
                       = 6Cr x12 2r (-y)r

Question-18

Prove that (x2 – yx)12, x ¹ 0

Solution:
Tr + 1 in (x2 -yx)12 = 12Cr (x2)12 - r (-yx)r
                         = 12Cr x24 – 2r (-1)r(y)r(x)r
                         = 12Cr x24 – r yr(-1)r

Question-19

Find the 4th term in the expansion of (x - 2y)12.

Solution:
4th term in (x - 2y)12 = T4 = T3 + 1
                              = 12C3(x)12 - 3(-2y)
3
     [Tr + 1 in (a + b)n = nCr an – r br]
                              = 12C3(x)9 (-2)3 (y)3
                              =
                              = -1760 x9 y
3

Question-20

Find the 13th term in the expansion of

Solution:
13th term in = T13 = T12 + 1
                                  = 18C12(9x)18 - 12
                                  = 18C12(9x)6

                                  = 18C12(9x)6
                                  = 18C12(9x)6
                                  = 18C12(32)6
                                  = 18C12(312)
                                  = 18C12 = 18564

Question-21

Find the 13th term in the expansion of

Solution:
The index of, is 7, which is an odd natural number.

So, Middle terms are and

= T4 = T3+1 = 7C3(3)7 – 3
                         = 7C3(3)4(x3)3(-1)(6-1)3
                         = 7C3(81)(x)9 (6)-3(-1)3
                         = =
= T5 = T4 + 1 = 7C4(3)7 – 4
                            = 7C4(3)3 (-1)4(x3)(6)-4
                            = 7C4(27) (x)12(6)-4
                            = (35)(27)(6)-4(x)12
                            = =

Question-22


Solution:
The index is 10, which is an even natural number.

Hence, Middle term = = T6 = T5+1
                           = 10C5(9y)5
                           = 10C5(9y)5
                           = 10C5(x)5
                           = 10C5(3)-5

                           = 10C5(3)-5
                           = 10C535 x5 y5 = (252)(243)x5 y
5
                           = 61236 x5y5

Question-23

In the expansions of (1 + a)m+n, using Binomial Theorem, prove that coefficients of am and an are equal.

Solution:
We have,

(1 + a)m+n = [m+nC0 + m+nC1a1 + m+nC2a2+…… m+nCr

ar + …… + m+nCm+n am+n

Coefficient of am = m+nCm =

Also the coefficient of an

= m+nCn =

Clearly, m+nCm = m+nCn

Question-24

The coefficients of the (r-1)th, rth and (r+1)th terms in the expansion of (x+1)n are in the ratio 1:3:5. Find both n and r.

Solution:
Coefficient of (r-1)th term = C(n, r-2)

Coefficient of rth term = C(n, r-1)

Coefficient of (r+1)th term = C(n, r)

Considering 1st and 2nd

 

3r - 3 = n - r + 2

n - 4r = -5 ----------(1)

Considering 2nd and 3rd

 

5r = 3n - 3r +3

3n - 8r = -3 ---------(2)

2(n - 4r = -5)

2n - 8r = -10 ---------(3) 

Subtract (3) from (2)

n = 7

Substitute n = 7 in (2)

We get r = 3

n = 7, r = 3

Question-25

Prove that the coefficient of xn in the expansion of (1+x)2n is twice the coefficient of xn in the expansion (1 + x)2n – 1.

Solution:
(1+x)2n = 2nC0+2nC1x1+2nC2x2 +…. + 2nCnxn

(1+x)2n-1 = 2n-1C0+2n-1C1x1 + 2n-1C2x2 +…. + 2n-1Cnxn

Coefficient of xn in (1+x)2n – 1 is (2n - 1Cn)

2nCn =

       =

       =

       = 2 …….(i)

Now 2n - 1Cn =

                  =

                  =

                  =

                  = …….(ii)

From (i) and (ii) we have

2nCn = 2. 2n-1Cn





Test Your Skills Now!
Take a Quiz now
Reviewer Name