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Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8
• Last Updated : 07 Apr, 2021

### Question 1. Find a, b, and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290, and 30375, respectively.

Solutions:

As, we know that (r+1)th term of (a+b)n is denoted by,

Tr+1 = nCr an-r br

Here, it is given that first three terms of the expansion are 729, 7290 and 30375.

When, T1 = 729, T2 = 7290 and T3 = 30375

T0+1 (r=0) = nC0 an-0 b0 = an = 729  …………………(1)

T1+1 (r=0) = nC1 an-1 b1 = nan-1 b = 7290  …………………(2)

T2+1 (r=0) = nC2 an-2 b2 an-2b2 = 30375  …………………(3)

Dividing (1) and (2), we get nan-1-n b = 10 = 10 ……………………….(I)

Now, dividing (3) and (2), we get    From (I), we can substitute  = 10 –  ………………….(II)

Substituting (II) in (I), we get = 10 = 10

n = n = 6

Substituting n = 6 in (1), we get

an = 729

a6 = 729

a = 3

Substituting a = 3 in (II), we get b = b = 5

Hence, a = 3, b = 5 and n = 6

### Question 2. Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Solutions:

As, we know that (r+1)th term of (a+b)n is denoted by,

Tr+1 = nCr an-r br

Here, a = 3 and b = ax and n = 9.

Tr+1 = 9Cr 39-r (ax)r

Tr+1 = 9Cr arxr

Tr+1 = 9Cr So, here if you want the power of x2 and x3. Then r=2 and r=3.

r = 2, T2+1 = r = 3, T3+1 Coefficient of x2 = Coefficient of x3      a = Hence, a = ### Question 3. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Solutions:

For getting the coefficient of x5, lets expand both the binomials for more clear understanding.

(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 = 7C07C1 (x) + 7C2 (x)27C3 (x)3 + 7C4 (x)47C5 (x)5 + 7C6 (x)67C7 (x)7

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

Now, the product will be seen as follows

(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)

Here, what we can see that x5 will be obtained when two terms will be multiplied of having sum of power of x as 5. Those two terms will be as follows:

So,

Coefficient of x5 = (1)(-21) + (12)(35) + (60)(-35) + (160)(21) + (240)(-7) + (192)(1)

Coefficient of x5 = -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x5 = -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x5 = 171

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

### [Hint write an = (a – b + b)n and expand]

Solutions:

To prove that (a – b) is a factor of (an – bn),

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

a can be written as = a – b + b

an = (a – b + b)n = [(a – b) + b]n

[(a – b) + b]n = nC0 (a – b)n + nC1 (a – b)n-1 b + ……….nCn-1(a – b)bn-1 + nCn bn

an = (a – b)n + n (a – b)n-1 b + ……….nCn-1(a – b)bn-1 + bn

Now, an – bn will be

an – bn = [(a – b)n + n (a – b)n-1 b + ……….nCn-1(a – b)bn-1 + bn] – bn

an – bn = (a – b)n + n (a – b)n-1 b + ……….nCn-1(a – b)bn-1

Taking (a-b) common, we have

an – bn = (a – b) [(a –b)n-1 + n (a – b)n-2 b + …… + nCn-1 bn-1]

an – bn = (a – b) k

Where k = [(a –b)n-1 + n (a – b)n-2 b + …… + nCn-1 bn-1] is a natural number

Hence, it is proved a – b is a factor of an – bn, where n is a positive integer

### Question 5. Evaluate (√3​+√2​)6−(√3​−√2​)6.

Solutions:

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

(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 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 a b5 + 6C6 b6

(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 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 a b5 + 6C6 b6]

(a + b)6 – (a – b)6 = 2[6C1 a5 b + 6C3 a3 b3 + 6C5 a b5]

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

### Question 6. Find the value of .

Solutions:

Using binomial theorem the expression (x+y)4 and (x – y)4, can be expanded as follows:

(x + y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4

(x – y)4 = 4C0 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4  y4

(x + y)4 + (x – y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 + [4C0 x44C1 x3 y + 4C2 x2 y24C3 x y3 + 4C4 y4]

(x + y)4 + (x – y)4 = 2[4C0 x4 + 4C2 x2 y2 + 4C4 y4]

Substituting x = a2 and y = , we get = 2[a8 + 6a4 (a2-1) + (a2-1)2]

= 2[a8 + 6a6 – 6a4 + (a4 + 1 – 2(a2)(1))]

= 2[a8 + 6a6 – 6a4 + a4 + 1 – 2a2]

= 2a8 + 12a6 – 10a4 – 4a2 + 2

### Question 7. Find an approximation of (0.99)5 using the first three terms of its expansion.

Solutions:

To make 0.99 in binomial form,

0.99 = 1 – 0.01

Now by applying binomial theorem, we get

(0. 99)5 = (1 – 0.01)5

Taking first three terms of its expansion, we have

= 5C0 (1)55C1 (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

Approximation of (0.99)5 = 0.951.

### Question 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is √6:1.

Solutions:

As, here it is said we have to calculate

(Fifth term from the beginning : Fifth term from the end) of the Binomial = Instead of taking fifth term from the end, lets reverse the binomial term and take it from beginning.

Fifth term from the end of binomial = Fifth term from the beginning Lets move further with this

As, we know that (r+1)th term of (a+b)n is denoted by,

Tr+1 = nCr an-r br

Fifth term from the beginning of T5 = T4+1 = nC4 T5 = nC4 T5 = nC4 T5 = nC4 …………………………….(1)

Now, Fifth term from the beginning of ,

T5 = T4+1 = nC4 T5 = nC4 T5 = nC4 T5 = nC4 T5 = nC4 ……………………….(2)

Now taking ratio of (1) and (2), which is equal to √6:1     n = n = 10

### Question 9. Expand using Binomial Theorem , x≠0.

Solutions:

Grouping in binomial form, we have Comparing it with (a+b)n,

a = , b = and n = 4 = 4C0 –  4C1 + 4C2 – 4C3 + 4C4    Now, lets get the value of and    = 3C0 (1)3 + 3C1 (1)2 + 3C2 (1) + 3C3   = 4C0 (1)4 + 4C1 (1)3 + 4C2 (1)2 + 4C3 (1) + 4C4  Now, substituting these values in the main equation, we get    ### Question 10. Find the expansion of (3x2– 2ax + 3a2)3 using binomial theorem.

Solutions:

Grouping (3x2– 2ax + 3a2)3 in binomial form, we have

[3x2 + (- 2ax + 3a2)]3

Comparing it with (a+b)n,

a = 3x2, b = -a (2x-3a) and n = 3

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

= 3C0 (3x2)3 +  3C1 (3x2)2 (-a (2x-3a)) +  3C2 (3x2) (-a (2x-3a))2 +  3C3 (-a (2x-3a))3

= 27x6 +  3 (9x4) (-a) (2x-3a) +  3 (3x2) (-a)2 (2x-3a)2 + (-a)3 (2x-3a)3

= 27x6 + (-54ax5 + 81a2x4) +  9a2x2 (2x-3a)2 – a3 (2x-3a)3

Now, lets get the value of (2x-3a)2 and (2x-3a)3.

(2x-3a)2 = (2x)2 + (3a)2 – 2(2x)(3a)

(2x-3a)2 = 4x2 + 9a2 -12xa

(2x-3a)3 = (2x)3 – (3a)3 – 3(2x)(3a)(2x-3a)

(2x-3a)3 = 8x3 – 27a3 – 36x2a +54xa2

Now, substituting these values in the main equation, we get

= 27x6 – 54ax5 + 81a2x4 +  9a2x2 (4x2 + 9a2 -12xa) – a3 (8x3 – 27a3 – 36x2a + 54xa2)

= 27x6 – 54ax5 + 81a2x4 + 36a2x4 + 81a4x2 -108x3a3 – (8a3x3 – 27a6 – 36x2a4 + 54xa5)

= 27x6 – 54ax5 + 117a2x4 + 81a4x2 -108x3a3 – 8a3x3 + 27a6 + 36x2a4 – 54xa5

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

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