# Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8

### 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,Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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T_{r+1}=^{n}C_{r}a^{n-r}b^{r}Here, it is given that first three terms of the expansion are 729, 7290 and 30375.

When, T

_{1 }= 729, T_{2}= 7290 and T_{3}= 30375T

_{0+1}(r=0) =^{n}C_{0}a^{n-0}b^{0}= a^{n }= 729…………………(1)T

_{1+1 }(r=0) =^{n}C_{1}a^{n-1}b^{1}= na^{n-1}b = 7290…………………(2)T

_{2+1}(r=0) =^{n}C_{2}a^{n-2}b^{2}= a^{n-2}b^{2 }= = 30375…………………(3)Dividing (1) and (2), we get

na

^{n-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 = 6Substituting n = 6 in (1), we get

a

^{n}= 729a

^{6}= 729

a = 3Substituting 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 x^{2} and x^{3} 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,

T_{r+1}=^{n}C_{r}a^{n-r}b^{r}Here, a = 3 and b = ax and n = 9.

T

_{r+1}=^{9}C_{r}3^{9-r}(ax)^{r}T

_{r+1}=^{9}C_{r}a^{r}x^{r}T

_{r+1}=^{9}C_{r}So, here if you want the power of x

^{2}and x^{3}. Then r=2 and r=3.r = 2, T

_{2+1 = }r = 3, T

_{3+1}=Coefficient of x

^{2}= Coefficient of x^{3}a =

Hence, a =

### Question 3. Find the coefficient of x^{5} in the product (1 + 2x)^{6} (1 – x)^{7} using binomial theorem.

**Solutions:**

For getting the coefficient of x

^{5}, lets expand both the binomials for more clear understanding.(1 + 2x)

^{6}=^{6}C_{0}+^{6}C_{1}(2x) +^{6}C_{2}(2x)^{2}+^{6}C_{3}(2x)^{3}+^{6}C_{4}(2x)^{4}+^{6}C_{5}(2x)^{5}+^{6}C_{6}(2x)^{6}= 1 + 6 (2x) + 15 (2x)

^{2}+ 20 (2x)^{3}+ 15 (2x)^{4}+ 6 (2x)^{5}+ (2x)^{6}= 1 + 12 x + 60x

^{2}+ 160 x^{3}+ 240 x^{4}+ 192 x^{5 }+ 64x^{6}(1 – x)

^{7}=^{7}C_{0}–^{7}C_{1}(x) +^{7}C_{2}(x)^{2}–^{7}C_{3}(x)^{3}+^{7}C_{4}(x)^{4}–^{7}C_{5}(x)^{5}+^{7}C_{6}(x)^{6}–^{7}C_{7}(x)^{7}= 1 – 7x + 21x

^{2}– 35x^{3}+ 35x^{4}– 21x^{5}+ 7x^{6}– x^{7}Now, the product will be seen as follows

(1 + 2x)

^{6}(1 – x)^{7}= (1 + 12 x + 60x^{2}+ 160 x^{3}+ 240 x^{4}+ 192 x^{5}+ 64x^{6}) (1 – 7x + 21x^{2}– 35x^{3}+ 35x^{4}– 21x^{5}+ 7x^{6}– x^{7})Here, what we can see that x

^{5}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:

First binomial Second binomial 1 ^{st}term6 ^{th}term2 ^{nd}term5 ^{th}term3 ^{rd}term4 ^{th}term4 ^{th}term3 ^{rd}term5 ^{th}term2 ^{nd}term6 ^{th}term1 ^{st}termSo,

Coefficient of x

^{5}= (1)(-21) + (12)(35) + (60)(-35) + (160)(21) + (240)(-7) + (192)(1)Coefficient of x

^{5 =}-21 + 420 – 2100 + 3360 – 1680 + 192Coefficient of x

^{5}= -21 + 420 – 2100 + 3360 – 1680 + 192Coefficient of x

^{5}= 171Hence, the coefficient of x

^{5}in the expression (1+2x)^{6}(1-x)^{7}is 171.

### Question 4. If a and b are distinct integers, prove that a – b is a factor of a^{n} – b^{n}, whenever n is a positive integer.

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

**Solutions:**

To prove that (a – b) is a factor of (a

^{n}– b^{n}),a

^{n}– b^{n}= k (a – b) where k is some natural number or constant.a can be written as = a – b + b

a

^{n}= (a – b + b)^{n}= [(a – b) + b]^{n}[(a – b) + b]

^{n}=^{n}C_{0}(a – b)^{n}+^{n}C_{1}(a – b)^{n-1 }b + ……….^{n}C_{n-1}(a – b)b^{n-1}+^{n}C_{n}b^{n}a

^{n}= (a – b)^{n}+ n (a – b)^{n-1}b + ……….^{n}C_{n-1}(a – b)b^{n-1 }+ b^{n}Now, a

^{n}– b^{n}will bea

^{n}– b^{n}= [(a – b)^{n}+ n (a – b)^{n-1}b + ……….^{n}C_{n-1}(a – b)b^{n-1}+ b^{n}] – b^{n}a

^{n}– b^{n}= (a – b)^{n }+ n (a – b)^{n-1}b + ……….^{n}C_{n-1}(a – b)b^{n-1}Taking (a-b) common, we have

a

^{n}– b^{n}= (a – b) [(a –b)^{n-1}+ n (a – b)^{n-2}b + …… +^{n}C_{n-1}b^{n-1}]a

^{n}– b^{n}= (a – b) kWhere k = [(a –b)

^{n-1}+ n (a – b)^{n-2}b + …… +^{n}C_{n-1}b^{n-1}] is a natural numberHence, it is proved

a – b is a factor of a, where n is a positive integer^{n}– b^{n}

### 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}=^{6}C0 a^{6}+^{6}C_{1}a^{5}b +^{6}C_{2}a^{4}b^{2}+^{6}C_{3}a^{3}b^{3}+^{6}C_{4}a^{2}b^{4}+^{6}C_{5}a b^{5}+^{6}C_{6}b^{6}(a – b)

^{6}=^{6}C_{0}a^{6}–^{6}C_{1}a^{5}b +^{6}C_{2 }a^{4}b^{2}–^{6}C_{3}a^{3}b^{3}+^{6}C_{4}a^{2}b^{4}–^{6}C_{5}a b^{5 }+^{6}C_{6}b^{6}Now adding them,

(a + b)

^{6}– (a – b)^{6}=^{6}C_{0}a^{6}+^{6}C_{1}a^{5}b +^{6}C_{2}a^{4}b^{2}+^{6}C_{3}a^{3}b^{3}+^{6}C_{4}a^{2}b^{4}+^{6}C_{5}a b^{5}+^{6}C_{6}b^{6}– [^{6}C_{0}a^{6}–^{6}C_{1}a^{5}b +^{6}C_{2}a^{4}b^{2}–^{6}C_{3}a^{3}b^{3}+^{6}C_{4}a^{2}b^{4}–^{6}C_{5}a b^{5}+^{6}C_{6}b^{6}]

(a + b)^{6}– (a – b)^{6}= 2[^{6}C_{1}a^{5}b +^{6}C_{3 }a^{3}b^{3}+^{6}C_{5}a b^{5}]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}=^{4}C_{0}x^{4}+^{4}C_{1}x^{3}y +^{4}C_{2}x^{2}y^{2 }+^{4}C_{3}x y^{3}+^{4}C_{4}y^{4}(x – y)

^{4}=^{4}C_{0}x^{4}–^{4}C_{1}x^{3}y +^{4}C_{2}x^{2}y^{2}–^{4}C_{3}x y^{3}+^{4}C_{4}y^{4}Now adding them,

(x + y)

^{4}+ (x – y)^{4}=^{4}C_{0}x^{4}+^{4}C_{1}x^{3}y +^{4}C_{2}x^{2}y^{2}+^{4}C_{3}x y^{3}+^{4}C_{4}y^{4}+ [^{4}C_{0}x^{4}–^{4}C_{1}x^{3}y +^{4}C_{2}x^{2}y^{2}–^{4}C_{3}x y^{3}+^{4}C_{4}y^{4}](x + y)

^{4}+ (x – y)^{4}= 2[^{4}C_{0}x^{4}+^{4}C_{2}x^{2}y^{2}+^{4}C_{4}y^{4}]Substituting x = a

^{2}and y = , we get= 2[a

^{8}+ 6a^{4}(a^{2}-1) + (a^{2}-1)^{2}]= 2[a

^{8}+ 6a^{6}– 6a^{4}+ (a^{4}+ 1 – 2(a^{2})(1))]= 2[a

^{8}+ 6a^{6}– 6a^{4}+ a^{4}+ 1 – 2a^{2}]= 2a

^{8}+ 12a^{6}– 10a^{4}– 4a^{2}+ 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

=

^{5}C_{0}(1)^{5}–^{5}C_{1}(1)^{4}(0.01) +^{5}C_{2}(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,

T_{r+1}=^{n}Cr a^{n-r}b^{r}Fifth term from the beginning of ,

T

_{5}= T_{4+1}=^{n}C_{4}T

_{5}=^{n}C_{4}T

_{5}=^{n}C^{4}T

_{5}=^{n}C_{4}…………………………….(1)Now, Fifth term from the beginning of ,

T

_{5}= T_{4+1}=^{n}C_{4}T

_{5}=^{n}C_{4}T

_{5}=^{n}C_{4}T

_{5}=^{n}C_{4}T

_{5}=^{n}C_{4}……………………….(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

=

^{4}C_{0}–^{4}C_{1}+^{4}C_{2}–^{4}C_{3}+^{4}C_{4}=

=

=

Now, lets get the value of and

=

^{3}C_{0}(1)^{3}+^{3}C_{1}(1)^{2}+^{3}C_{2}(1) +^{3}C_{3}=

^{4}C_{0}(1)^{4}+^{4}C_{1}(1)^{3}+^{4}C_{2 }(1)^{2}+^{4}C_{3}(1) +^{4}C_{4}Now, substituting these values in the main equation, we get

=

=

=

=

### Question 10. Find the expansion of (3x^{2}– 2ax + 3a^{2})^{3} using binomial theorem.

**Solutions:**

Grouping (3x

^{2}– 2ax + 3a^{2})^{3}in binomial form, we have[3x

^{2 }+ (- 2ax + 3a^{2})]^{3}Comparing it with (a+b)

^{n},a = 3x

^{2}, b = -a (2x-3a) and n = 3[3x

^{2}+ (-a (2x-3a))]^{3}=

^{3}C_{0}(3x^{2})^{3}+^{3}C_{1 }(3x^{2})^{2 }(-a (2x-3a)) +^{3}C_{2}(3x^{2}) (-a (2x-3a))^{2 }+^{3}C_{3 }(-a (2x-3a))^{3}= 27x

^{6}+ 3 (9x^{4}) (-a) (2x-3a) + 3 (3x^{2}) (-a)^{2 }(2x-3a)^{2}+ (-a)^{3}(2x-3a)^{3}= 27x

^{6}+ (-54ax^{5}+ 81a^{2}x^{4}) + 9a^{2}x^{2}(2x-3a)^{2}– a^{3}(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 }= 4x^{2}+ 9a^{2}-12xa(2x-3a)

^{3}= (2x)^{3}– (3a)^{3}– 3(2x)(3a)(2x-3a)(2x-3a)

^{3}= 8x^{3}– 27a^{3}– 36x^{2}a +54xa^{2}Now, substituting these values in the main equation, we get

= 27x

^{6}– 54ax^{5}+ 81a^{2}x^{4}+ 9a^{2}x^{2}(4x^{2}+ 9a^{2}-12xa) – a^{3}(8x^{3 }– 27a^{3}– 36x^{2}a + 54xa^{2})= 27x

^{6}– 54ax^{5}+ 81a^{2}x^{4}+ 36a^{2}x^{4}+ 81a^{4}x^{2}-108x^{3}a^{3}– (8a^{3}x^{3}– 27a^{6}– 36x^{2}a^{4 }+ 54xa^{5)}= 27x

^{6 }– 54ax^{5}+ 117a^{2}x^{4}+ 81a^{4}x^{2}-108x^{3}a^{3}– 8a^{3}x^{3}+ 27a^{6}+ 36x^{2}a^{4}– 54xa^{5}= 27x

^{6}– 54ax^{5}+ 117a^{2}x^{4}– 116a^{3}x^{3}+ 117a^{4}x^{2}– 54a^{5}x + 27a^{6}