NCERT Class 11 Maths Chapter 7 Binomial Theorem Solutions

 

From the binomial theorem, the given equation can be expanded as

 

From the binomial theorem, the given equation can be expanded as

 

From binomial theorem expansion, we can write as

(1 – 2x)5

= 5Co (1)5 – 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 – 5C3 (1)2 (2x)3 + 5C4 (1)1 (2x)4 – 5C5 (2x)5

= 1 – 5 (2x) + 10 (4x)2 – 10 (8x3) + 5 ( 16 x4) – (32 x5)

= 1 – 10x + 40x2 – 80x3 + 80x4– 32x5

 

From the binomial theorem, the given equation can be expanded as

 

From the binomial theorem, the given equation can be expanded as

 

Given (96)3

96 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.

The given question can be written as 96 = 100 – 4

(96)3 = (100 – 4)3

= 3C0 (100)3 – 3C1 (100)2 (4) – 3C2 (100) (4)2– 3C3 (4)3

= (100)3 – 3 (100)2 (4) + 3 (100) (4)2 – (4)3

= 1000000 – 120000 + 4800 – 64

= 884736

 

Given (102)5

102 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.

The given question can be written as 102 = 100 + 2

(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

= (100)5 + 5 (100)4 (2) + 10 (100)3 (2)2 + 5 (100) (2)3 + 5 (100) (2)4 + (2)5

= 1000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

 

Given (101)4

101 can be expressed as the sum or difference of two numbers, and then the binomial theorem can be applied.

The given question can be written as 101 = 100 + 1

(101)4 = (100 + 1)4

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

= (100)4 + 4 (100)3 + 6 (100)2 + 4 (100) + (1)4

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

 

 

Given (99)5

99 can be written as the sum or difference of two numbers then the binomial theorem can be applied.

The given question can be written as 99 = 100 -1

(99)5 = (100 – 1)5

= 5C0 (100)5 – 5C1 (100)4 (1) + 5C2 (100)3 (1)2 – 5C3 (100)2 (1)3 + 5C4 (100) (1)4 – 5C5 (1)5

= (100)5 – 5 (100)4 + 10 (100)3 – 10 (100)2 + 5 (100) – 1

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

 

 

Given (99)5

99 can be written as the sum or difference of two numbers then the binomial theorem can be applied.

The given question can be written as 99 = 100 -1

(99)5 = (100 – 1)5

= 5C0 (100)5 – 5C1 (100)4 (1) + 5C2 (100)3 (1)2 – 5C3 (100)2 (1)3 + 5C4 (100) (1)4 – 5C5 (1)5

= (100)5 – 5 (100)4 + 10 (100)3 – 10 (100)2 + 5 (100) – 1

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

 

By splitting the given 1.1 and then applying the binomial theorem, the first few terms of (1.1)10000 can be obtained as

(1.1)10000 = (1 + 0.1)10000

= (1 + 0.1)10000 C1 (1.1) + other positive terms

= 1 + 10000 × 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)10000 > 1000

 

Using the binomial theorem, the expression (a + b)4 and (a – b)4 can be expanded

(a + b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4

(a – b)4 = 4C0 a4 – 4C1 a3 b + 4C2 a2 b2 – 4C3 a b3 + 4C4 b4

Now (a + b)4 – (a – b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4 – [4C0 a4 – 4C1 a3 b + 4C2 a2 b2 – 4C3 a b3 + 4C4 b4]

= 2 (4C1 a3 b + 4C3 a b3)

= 2 (4a3 b + 4ab3)

= 8ab (a2 + b2)

Now by substituting a = √3 and b = √2, we get

(√3 + √2)4 – (√3 – √2)4 = 8 (√3) (√2) {(√3)2 + (√2)2}

= 8 (√6) (3 + 2)

= 40 √6

 

Using binomial theorem, the expressions (x + 1)6 and (x – 1)6 can be expressed as

(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6

(x – 1)6 = 6C0 x6 – 6C1 x5 + 6C2 x4 – 6C3 x3 + 6C4 x2 – 6C5 x + 6C6

Now, (x + 1)6 – (x – 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6 – [6C0 x6 – 6C1 x5 + 6C2 x4 – 6C3 x3 + 6C4 x2 – 6C5 x + 6C6]

= 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6]

= 2 [x6 + 15x4 + 15x2 + 1]

Now by substituting x = √2, we get

(√2 + 1)6 – (√2 – 1)6 = 2 [(√2)6 + 15(√2)4 + 15(√2)2 + 1]

= 2 (8 + 15 × 4 + 15 × 2 + 1)

= 2 (8 + 60 + 30 + 1)

= 2 (99)

= 198

 

In order to show that 9n+1 – 8n – 9 is divisible by 64, it has to be shown that 9n+1 – 8n – 9 = 64 k, where k is some natural number.

Using the binomial theorem,

(1 + a)m = mC0 + mC1 a + mC2 a2 + …. + m C m am

For a = 8 and m = n + 1 we get

(1 + 8)n+1 = n+1C0 + n+1C1 (8) + n+1C2 (8)2 + …. + n+1 C n+1 (8)n+1

9n+1 = 1 + (n + 1) 8 + 82 [n+1C2 + n+1C3 (8) + …. + n+1 C n+1 (8)n-1]

9n+1 = 9 + 8n + 64 [n+1C2 + n+1C3 (8) + …. + n+1 C n+1 (8)n-1]

9n+1 – 8n – 9 = 64 k

Where k = [n+1C2 + n+1C3 (8) + …. + n+1 C n+1 (8)n-1] is a natural number

Thus, 9n+1 – 8n – 9 is divisible by 64 whenever n is a positive integer.

Hence proved.

 

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here x5 is the Tr+1 term so a= x, b = 3 and n =8

Tr+1 = 8Cr x8-r 3r…………… (i)

To find out x5

We have to equate x5= x8-r

⇒ r= 3

Putting the value of r in (I), we get

= 1512 x5

Hence the coefficient of x5= 1512.

 

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here a = a, b = -2b & n =12

Substituting the values, we get

Tr+1 = 12Cr a12-r (-2b)r………. (i)

To find a5

We equate a12-r =a5

r = 7

Putting r = 7 in (i)

T8 = 12C7 a5 (-2b)7

= -101376 a5 b7

Hence, the coefficient of a5b7= -101376.

 

The general term Tr+1 in the binomial expansion is given by

Tr+1 = n C r an-r br…….. (i)

Here, a = x2 , n = 6 and b = -y

Putting values in (i)

Tr+1 = 6Cr x 2(6-r) (-1)r yr

= -1r 6cr .x12 – 2r. yr

 

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here n = 12, a= x2 and b = -y x

Substituting the values, we get

Tn+1 =12Cr × x2(12-r) (-1)r yr xr

= -1r 12cr .x24 –2r. yr

 

The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br

Here, a= x, n =12, r= 3 and b = -2y

By substituting the values, we get

T4 = 12C3 x9 (-2y)3

= -1760 x9 y3

 

We know that the general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

Tr+1 = m+n Cr 1m+n-r ar

= m+n Cr ar…………. (i)

Now, we have that the general term for the expression is,

Tr+1 =  m+n Cr ar

Now, for coefficient of am

Tm+1 =  m+n Cm am

Hence, for the coefficient of am, the value of r = m

So, the coefficient is m+n C m

Similarly, the coefficient of an is m+n C n

 

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here, the binomial is (1+x)n with a = 1 , b = x and n = n

The (r+1)th term is given by

T(r+1) = nCr 1n-r xr

T(r+1) = nCr xr

The coefficient of (r+1)th term is nCr

The rth term is given by (r-1)th term

T(r+1-1) = nCr-1 xr-1

Tr = nCr-1 xr-1

∴ the coefficient of rth term is nCr-1

For (r-1)th term, we will take (r-2)th term

Tr-2+1 = nCr-2 xr-2

Tr-1 = nCr-2 xr-2

∴ the coefficient of (r-1)th term is nCr-2

Given that the coefficient of (r-1)th, rth and r+1th term are in ratio 1:3:5

Therefore,

⇒ 5r = 3n – 3r + 3

⇒ 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equations 2 and 3

2n -8r +10 =0

-3n – 8r – 3 =0

⇒ -n = -7

n =7 and r = 3

 

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

The general term for binomial (1+x)2n is

Tr+1 = 2nCr xr …………………..1

To find the coefficient of xn

r = n

Tn+1 = 2nCn xn

The coefficient of xn = 2nCn

The general term for binomial (1+x)2n-1 is

Tr+1 = 2n-1Cr xr

To find the coefficient of xn

Putting n = r

Tr+1 = 2n-1Cr xn

The coefficient of xn = 2n-1Cn

We have to prove

Coefficient of xn in (1+x)2n = 2 coefficient of xn in (1+x)2n-1

Consider LHS = 2nCn

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here, a = 1, b = x and n = m

Putting the value

Tr+1 = m Cr 1m-r xr

= m Cr xr

We need the coefficient of x2

∴ putting r = 2

T2+1 = mC2 x2

The coefficient of x2 = mC2

Given that coefficient of x2 = mC2 = 6

⇒ m (m – 1) = 12

⇒ m2– m – 12 =0

⇒ m2– 4m + 3m – 12 =0

⇒ m (m – 4) + 3 (m – 4) = 0

⇒ (m+3) (m – 4) = 0

⇒ m = – 3, 4

We need the positive value of m, so m = 4

 

We know that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = nCr an-t br

The first three terms of the expansion are given as 729, 7290 and 30375, respectively. Then we have,

T1 = nC0 an-0 b0 = an = 729….. 1

T2 = nC1 an-1 b1 = nan-1 b = 7290…. 2

T3 = nC2 an-2 b2 = {n (n -1)/2 }an-2 b2 = 30375……3

Dividing 2 by 1, we get

nan−1ban=7290729nba=10

Dividing 3 by 2, we get

n(n−1)an−2b22nan−1b=303757290(n−1)b2a=303757290(n−1)ba=303757290×2=253(nb)a−ba=25310−ba=253ba=10–253=53

From 4 and 5, we have

n. 5/3 = 10

n = 6

Substituting n = 6 in 1, we get

a6 = 729

a = 3

From 5, we have, b/3 = 5/3

b = 5

Thus a = 3, b = 5 and n = 76

 

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

= 1 + 6 (2x) + 15 (2x)2 + 20 (2x)3 + 15 (2x)4 + 6 (2x)5 + (2x)6

= 1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6

(1 – x)7 = 7C0 – 7C1 (x) + 7C2 (x)2 – 7C3 (x)3 + 7C4 (x)4 – 7C5 (x)5 + 7C6 (x)6 – 7C7 (x)7

= 1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7

(1 + 2x)6 (1 – x)7 = (1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6) (1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7)

192 – 21 = 171

Thus, the coefficient of x5 in the expression (1+2x)6(1-x)7 is 171.

 

In order to prove that (a – b) is a factor of (an – bn), it has to be proved that

an – bn = k (a – b) where k is some natural number.

a can be written as a = a – b + b

an = (a – b + b)n = [(a – b) + b]n

= nC0 (a – b)n + nC1 (a – b)n-1 b + …… + n C n bn

an – bn = (a – b) [(a –b)n-1 + nC1 (a – b)n-1 b + …… + n C n bn]

an – bn = (a – b) k

Where k = [(a –b)n-1 + nC1 (a – b)n-1 b + …… + n C n bn] is a natural number

This shows that (a – b) is a factor of (an – bn), where n is a positive integer.

 

Using the binomial theorem, the expression (a + b)6 and (a – b)6 can be expanded

(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6

(a – b)6 = 6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 a b5 + 6C6 b6

Now (a + b)6 – (a – b)6 =6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6 – [6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 a b5 + 6C6 b6]

Now by substituting a = √3 and b = √2, we get

(√3 + √2)6 – (√3 – √2)6 = 2 [6 (√3)5 (√2) + 20 (√3)3 (√2)3 + 6 (√3) (√2)5]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

 

0.99 can be written as

0.99 = 1 – 0.01

Now by applying the binomial theorem, we get

(o. 99)5 = (1 – 0.01)5

= 5C0 (1)5 – 5C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2

= 1 – 5 (0.01) + 10 (0.01)2

= 1 – 0.05 + 0.001

= 0.951

 

Using the binomial theorem, the given expression can be expanded as

Again by using the binomial theorem to expand the above terms, we get

From equations 1, 2 and 3, we get

We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3

Putting a = 3x2 & b = -a (2x-3a), we get

[3x2 + (-a (2x-3a))]3

= (3x2)3+3(3x2)2(-a (2x-3a)) + 3(3x2) (-a (2x-3a))2 + (-a (2x-3a))3

= 27x6 – 27ax4 (2x-3a) + 9a2x2 (2x-3a)2 – a3(2x-3a)3

= 27x6 – 54ax5 + 81a2x4 + 9a2x2 (4x2-12ax+9a2) – a3 [(2x)3 – (3a)3 – 3(2x)2(3a) + 3(2x)(3a)2]

= 27x6 – 54ax5 + 81a2x4 + 36a2x4 – 108a3x3 + 81a4x2 – 8a3x3 + 27a6 + 36a4x2 – 54a5x

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6

Thus, (3x2 – 2ax + 3a2)3

= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6