RBSE Solutions for Class 11 Maths Chapter 8 – Binomial theorem are available here. The important questions and solutions of Chapter 8, available at BYJU’S, contain detailed explanations to every problem, making a student’s life easy. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions.

Chapter 8 of the RBSE Class 11 Maths will help the students to solve problems related to the binomial expression, binomial theorem for a positive index, important terms of the binomial theorem, general term in binomial expression, properties of binomial coefficients, binomial theorem for a rational index, important expansions, applications of the binomial theorem, the sum of series by binomial theorem. After referring to these important questions and solutions, you can refer to 2020-2021 solutions for the chapter here.

### RBSE Maths Chapter 8: Miscellaneous Questions and Solutions

**Question 1: The number of terms in the expansion of ((a / x) + bx) ^{12} are**

(A) 11

(B) 13

(C) 10

(D) 14

**Solution:**

The terms of the right-hand side of the expansion of (x + a)^{n }are finite for the positive values of x and the number of terms is (n + 1).

So, the value of n is 12 in the given expression.

Hence, number of total terms = 12 + 1 = 13.

Hence, option (B) is correct.

**Question 2: The 7 ^{th} term in the expansion of ((1 / 2) + a)^{8} is**

(A) ^{8}C_{7} (1 / 2) a^{7}

(B) ^{8}C_{7} (1 / 2) a

(C) ^{8}C_{6} (1 / 2)^{2} (a)^{6}

(D) ^{8}C_{6} (1 / 2)^{6} a^{2}

**Solution: **

The 7^{th }term in the expansion of ((1 / 2) + a)^{8} is

T_{7} = T_{6+1}

= ^{8}C_{6} (1 / 2)^{8-6} a^{6}

= ^{8}C_{6} (1 / 2)^{2} (a)^{6}

Hence, option (C) is correct.

**Question 3: If in the expansion of (1 + x) ^{18}, coefficients of (2r + 4)^{th} and (r – 2)^{th} terms are equal then value of r is: **

(A) 5

(B) 6

(C) 7

(D) 8

**Solution:**

In the expansion of (1 + x)^{18} coefficients of (2r + 4)^{th} and (r – 2)^{th} terms are ^{18}C_{2r+3} and ^{18}C_{r–3} respectively.

These are equal, then, ^{18}C_{2r+3 }= ^{18}C_{r–3}

⇒ 2r + 3 = r – 3

2r + 3 = 18 – (r – 3)

For using the statement if

^{n}C_{r} = ^{n}C_{p} or ^{n}C_{n-p}

⇒ 2r – r = -3 – 3

2r + 3 = 18 – r + 3

⇒ r = -6 or r = 6 r is a natural number, so r = – 6 is not possible, then x = 6.

Hence, option (B) is correct.

**Question 4: If in the expansion of (a + b) ^{n} and (a + b)^{n+3} ratio of 2^{nd} and 3^{rd}, 3^{rd} and 4^{th} terms are equal, then value of n is : **

(A) 5

(B) 6

(C) 3

(D) 4

**Solution:**

In the expansion of (a + b)^{n}

Hence, option (A) is correct.

**Question 5: If in the expansion of (1 + x) ^{2n}, coefficient of 3^{rd} and (r + 2)^{th} term are equal, then : **

(A) n = 2r

(B) n = 2r – 1

(C) n = 2r + 1

(D) n = r + 1** **

**Solution:**

In the expansion of (1 + x)^{2n}, 3^{rd} term = 2 ^{n}C_{3 –1} and (r + 2)^{th} term = ^{2n}C_{r+1}

According to the question, ^{2n}C_{3r–1} = ^{2n}C_{r+1}

Now, 3r – 1 = r + 1

3r – 1 = 2n – r – 1

⇒ 3r – r – I + 1

3r + r = 2n – 1 + 1

⇒ 2r = 2 or 4r = 2n

⇒ r =1

2r = n

Hence, option (A) is correct.

**Question 6: Find the value of ^{30}C_{1} + ^{30}C_{2} + ^{30}C_{3} +… + ^{30}C_{30}. **

**Solution:**

(1 + x)^{n} = ^{n}C_{0} + ^{n}C_{1}x + ^{n}C_{2}x^{2} + ^{n}C_{3}x^{3} + ….

Putting x = 1

(1 + 1)^{n} = ^{n}C_{0} + ^{n}C_{1}x + ^{n}C_{2}x^{2} + ^{n}C_{3}x^{3} + …….. ^{n}C_{n} or

2^{n} = ^{n}C_{0} + ^{n}C_{1}x + ^{n}C_{2}x^{2} + ^{n}C_{3}x^{3} + …….. ^{n}C_{n}

Here, putting n = 30.

2^{30} = 1 + ^{30}C_{1} + ^{30}C_{2} + ^{30}C_{3} +… + ^{30}C_{30}

⇒ 1+ ^{30}C_{1} + ^{30}C_{2} + ^{30}C_{3} +… + ^{30}C_{30} = 2^{30}

⇒ ^{30}C_{1} + ^{30}C_{2} + ^{30}C_{3} +… + ^{30}C_{30} = 2^{30} – 1

**Question 7: Find the middle term in the expansion of [(a / x) + (x / a)] ^{10}.**

**Solution:**

Here, the middle term is [(10 / 2) + 1]^{th} term = 6^{th} term.

T_{6} = T_{5+1}

= ^{10}C_{5} (a / x)^{10-5} (x / a)^{5}

= ^{10}C_{5} (a / x)^{5} (x / a)^{5}

= 252 * (1)

= 252

### RBSE Maths Chapter 8: Exercise 8.1 Textbook Important Questions and Solutions

**Question 1: Expand each expression in the following questions.**

**[i] (2 – x) ^{3}**

**[ii] (3x + 2y) ^{4}**

**[iii] [√(x / a) – √9 / x] ^{6}**

**Solution: **

**[i] (2 – x) ^{3}**

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

= 1 * 8 + 3 * 4 * (-x) + 3 * 2 * x^{2} + 1 * (-x)^{3}

= 8 – 12x + 6x^{2} – x^{3}

**[ii] (3x + 2y) ^{4} **

**[iii] [√(x / a) – √9 / x] ^{6}**

**Question 2: Using the binomial theorem, find the value of (1.1) ^{6}.**

**Solution: **

**(1.1) ^{6} **

= (1 + 0.1)^{6}

= ^{6}C_{0}(1)^{6} (0.1)^{0} + ^{6}C_{1} (1)^{5} (0.1)^{1} + ^{6}C_{2} (1)^{4} (0.1)^{2} + ^{6}C_{3} (1)^{3} (0.3)^{3} + ^{6}C_{4} (1)^{2} (0.1)^{4} + ^{6}C_{5} (1)^{1} (0.1)^{5} + ^{6}C_{6} (1)^{0} (0.1)^{6}

= (1 * 1 * 1) + (6 * 1 * 0.1) + {15 * 1 * (0.1)^{2}} + {20 * 1 * (0.1)^{3}} + {15 * 1 * (0.1)^{4}} + {6 * 1 * (0.1)^{5}} + {1 * 1 * (0.1)^{6}}

= 1 + 6 * 0.1 + 15 * (0.1)^{2} + 20 * (0.1)^{3} + 15 * (0.1)^{4} + 6 * (0.1)^{5} + (0.1)^{6}

= 1 + 0.6 + 15 * 0.01 + 20 * 0.001 + 15 * 0.0001 + 6 * 0.00001 + 0.000001

= 1 + 0.6 + 0.15 + 0.020 + 0.0015 + 0.00006 + 0.000001

= 1.771561

### RBSE Maths Chapter 8: Exercise 8.2 Textbook Important Questions and Solutions

**Question 1: In the following expansions, find the terms as stated.**

**[i] 5 ^{th} term of (a + 2x^{3})^{17}**

**[ii] 9 ^{th} term of ([x / y] – [3y / x^{2}])^{17}**

**[iii] 6 ^{th} term of ([2 / √x] – [x^{2} / 2])^{9}**

**Solution:**

**[i] 5 ^{th} term of (a + 2x^{3})^{17}**

The r^{th} term in an expansion of (a + x)^{n} is given by ^{n}C_{r−1} a^{n+1−r }x^{r−1}.

Similarly, the fifth term in the expansion of (a + 2x^{3})^{17} is given by

= ^{17}C_{4} a^{13 }(2x)^{4}

= [2380] * 16 a^{13 } * x^{12}

= 38080 a^{13 }x^{12}

**[ii] 9 ^{th} term of ([x / y] – [3y / x^{2}])^{17}**

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

a = [x / y], b = [3y / x^{2}], n = 12, r = 8

9^{th} term = T_{9} = T_{8+1}

T_{9} = ^{12}C_{8} [x / y]^{12-8} [3y / x^{2}]^{8}

= 495 * [x / y]^{4 }[3y / x^{2}]^{8}

= 495 * 6561 * x^{4-16} y^{-4+8}

= [3247695 / x^{12}] y^{4}

**[iii] 6 ^{th} term of ([2 / √x] – [x^{2} / 2])^{9}**

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

a = [2 / √x], b = [- x^{2} / 2], n = 9, r = 5

6^{th} term = T_{6} = T_{5+1}

T_{6} = ^{9}C_{5} [2 / √x]^{9-5} [- x^{2} / 2]^{5}

= 126 * [2 / √x]^{4 }[- x^{2} / 2]^{5}

= – [2016 / 32] * x^{10-[4/2]}

= – [63] x^{8}

**Question 2: Find the coefficient of the following.**

**(i) x ^{-7} in the expansion of [ax – (1 / bx^{2})]^{8}**

**(ii) x ^{4} in the expansion of [(x^{4} + [1 / x^{3}]]^{15}**

**(iii) x ^{6} in the expansion (a – bx^{2})^{10} **

**Solution:**

**(i) x ^{-7} in the expansion of [ax – (1 / bx^{2})]^{8}**

^{2})]

(r + 1)^{th} term of [ax – (1 / bx^{2})]^{8}

T_{r+1} = ^{8}C_{r} (ax)^{8-r} [-1 / bx^{2}]^{r}

T_{r+1} = (-1)^{r} ^{8}C_{r } a^{8-r} x^{8-r} [1 / b^{r} x^{2r}]

T_{r+1} = (-1)^{r } ^{8}C_{r} [a^{8-r} / b^{r}] x^{8-r-2r}

T_{r+1} = (-1)^{r} ^{8}C_{r }[a^{8-r} / b^{r}] x^{8-3r}

In this term for coefficient of x^{-7 }power of x should be

-7 = 8 – 3r = – 7

Hence,

8 – 3r = – 7

⇒ -3r = – 7 – 8 = – 15

r = 5

Hence, coefficient of x^{-7} in the expansion is

(-1)^{5} ^{8}C_{5 }[a^{8-5} / b^{5}] x^{8-3(5)}

= [- 56a^{3} / b^{5}]

**(ii) x ^{4} in the expansion of [(x^{4} + [1 / x^{3}]]^{15}**

^{4}+ [1 / x

^{3}]]

(r + 1)^{th} term of [(x^{4} + [1 / x^{3}]]^{15}

T_{r+1} = ^{15}C_{4} (x^{4})^{15-r} [1 / x^{3}]^{r}

T_{r+1} = ^{15}C_{r } x^{60-4r} [1 / x^{3}]^{r}

T_{r+1} = ^{15}C_{r } x^{60-4r-3r}

T_{r+1} = ^{15}C_{r } x^{60-7r}

In this term for coefficient of x^{4 }power of x should be

4 = 60 – 7r = -56

Hence,

8 – 3r = – 7

r = 8

Hence, the coefficient of x^{4} in the expansion is 6435.

**(iii) x ^{6} in the expansion (a – bx^{2})^{10}**

(a – bx^{2})

(r + 1)^{th} term of (a – bx^{2})^{10}

T_{r+1} = ^{10}C_{r} (a)^{10-r} [– bx^{2}]^{r}

T_{r+1} = ^{10}C_{r } a^{10-r} [-1]^{r} b^{r} x^{2r}

T_{r+1} = (-1)^{r} ^{10}C_{r } a^{10-r }b^{r} x^{2r}

In this term for coefficient of x^{6 }power of x should be

2r = 6

r = 3

Hence, coefficient of x^{6} in the expansion is

(-1)^{3} ^{10}C_{3 }[a^{10-3} b^{3}]

= [- 120a^{7} b^{3}]

**Question 3: In the following expansions, find the term independent of x.**

**[i] [x – (1 / x ^{2})]^{12}**

**[ii] [√x – (3 / x ^{2})]^{10}**

**[iii] [(√x / 3] + (3 / 2x ^{2})]^{10}**

**[iv] [x ^{2} – (1 / x^{2})]^{10}**

**Solution:**

**[i] [x – (1 / x ^{2})]^{12}**

(r + 1)^{th} term of [x – (1 / x^{2})]^{12}

T_{r+1} = ^{12}C_{r} (x)^{12-r} [(1 / x^{2})]^{r}

T_{r+1} = (-1)^{r} ^{12}C_{r } x^{12-r-2r}

T_{r+1} = (-1)^{r} ^{12}C_{r } x^{12-3r }

For the term independent of x,

x^{12-3r} = x^{0}

12 – 3r = 0

12 = 3r

12 / 3 = r

r = 4

Hence, the term independent of x is

(-1)^{4} ^{12}C_{4 }

= 495

**[ii] [√x – (3 / x ^{2})]^{10}**

(r + 1)^{th} term of [√x – (3 / x^{2})]^{10}

T_{r+1} = ^{10}C_{r} (√x)^{10-r} [- (3 / x^{2})]^{r}

T_{r+1} = ^{10}C_{r } x^{[10-r]/2} (-1)^{r} 3^{r} x^{-2r}

T_{r+1} = (-1)^{r} ^{10}C_{r }3^{r} x^{[10-r]/2 – [2r] }

For the term independent of x,

x^{[10-r]/2 – [2r]} = x^{0}

{[10 – r] / [2] – [2r]} = 0

10 – r – 4r = 0

10 – 5r = 0

r = 2

Hence, the term independent of x is

(-1)^{2} ^{10}C_{2} 3^{2}_{ }

= 45 * 9

= 405

**[iii] [(√x / 3] + (3 / 2x ^{2})]^{10}**

(r + 1)^{th} term of [(√x / 3)] + (3 / 2x^{2})]^{10}

T_{r+1} = ^{10}C_{r} (√x / 3)^{10-r} (3 / 2x^{2})^{r}

T_{r+1} = ^{10}C_{r } (x / 3)^{[5-r/2]} (3 / 2)^{r }x^{-2r}

T_{r+1} = ^{10}C_{r }(1 / 3)^{5-r/2} (3 / 2)^{r} x^{5-(r/2)-2r }

(r + 1)^{th} term is constant, that is the power of x should be 0.

5 – (r / 2) – 2r = 0

5 = 5r / 2

r = 2

The constant term = ^{10}C_{2} (1 / 3)^{5-(2/2)} (3 / 2)^{2} x^{5-1-4}

= ^{ 10}C_{2}^{ }(1 / 3)^{4} (3 / 2)^{2}

= 5 / 4

**[iv] [x ^{2} – (1 / x^{2})]^{10}**

(r + 1)^{th} term of [x^{2} – (1 / x^{2})]^{10}

T_{r+1} = ^{10}C_{r} (x^{2})^{10-r} [- (1 / x^{2})]^{r}

T_{r+1} = (-1)^{r 10}C_{r } x^{20-2r-2r}

T_{r+1} = (-1)^{r} ^{10}C_{r }x^{[20-4r] }

For the term independent of x,

x^{[20-4r]} = x^{0}

20 – 4r = 0

20 = 4r

20 / 4 = r

5 = r

Hence, the term independent of x is

(-1)^{5} ^{10}C_{5}

= -252

**Question 4: Find the middle term in the following expansions.**

**[i] ([x / 2] + 2y) ^{6}**

**[ii] [3a – (a ^{3} / 6)]^{9}**

**[iii] (x + (1 / x)) ^{2n}**

**[iv] [3x – (2 / x ^{2})]^{15}**

**Solution:**

**[i] ([x / 2] + 2y) ^{6}**

The middle term in the expansion of ([x / 2] + 2y)^{6}. In the expansion of ([x / 2] + 2y)^{6}, the middle term will be T_{[n/2 + 1]}^{th }term where n = 6 which is an even number.

The middle term T_{[n/2 + 1]} = T_{4}

T_{4} = T_{3+1}

= ^{6}C_{3} [x / 2]^{6-3} (2y)^{3}

= 20 * [x^{3} / 2^{3}] * 2^{3} * y^{3}

= 20 x^{3} y^{3}

**[ii] [3a – (a ^{3} / 6)]^{9}**

Here n = 9, which is an odd number.

The middle term is {[(n + 1) / 2] + [1]}^{th} term and {(n + 1) / 2}^{th} term.

Here the middle term is 5^{th} and 6^{th} term.

Now, T_{r+1} = ^{9}C_{r} (3a)^{9-r} (-a^{3} / 6)^{r}

Put r = 4,

T_{5} = ^{9}C_{4} (3a)^{9-4} (-a^{3} / 6)^{4}

= 126 * 3^{5} * a^{5} * [a^{12} / 2^{4} 3^{4}]

= [189 / 8] a^{17}

Put r = 5,

T_{6} = ^{9}C_{5} (3a)^{9-5} (-a^{3} / 6)^{5}

= -126 / 96 [a^{19}]

= [-21 / 16] a^{19}

The two middle terms are [189 / 8] a^{17} and [-21 / 16] a^{19}.

**[iii] (x + (1 / x)) ^{2n}**

Here n = 2n, which is an even number.

The middle term is {[(2n) / 2] + [1]}^{th} term and {(n + 1)}^{th} term.

Now, T_{r+1} = ^{2n}C_{r} (x)^{2n-r} (1 / x)^{r}

Put r = n,

T_{n+1} = ^{2n}C_{n} (x)^{2n-n} (1 / x^{n})

= {(2n)! / [n! * n!]} * x^{n }* (1 / x^{n})

= (2n)! / (n!)^{2}

The middle term is (2n)! / (n!)^{2}.

**[iv] [3x – (2 / x ^{2})]^{15}**

Here n = 15, which is an odd number.

The middle term is {[(n + 1) / 2] + [1]}^{th} term and {(n + 1) / 2}^{th} term.

Here the middle term is 8^{th} and 9^{th} term.

Now, T_{r+1} = ^{15}C_{r} (3x)^{15-r} (-2 / x^{2})^{r}

Put r = 7,

T_{8} = ^{15}C_{7} (3x)^{15-7} (-2 / x^{2})^{7}

= [6435 * 3^{8} * (-1)^{7} * 2^{7} * x^{8}] / x^{14}

= [-6435 * 3^{8} * 2^{7}] / x^{6}

Put r = 8,

T_{9} = ^{15}C_{8} (3x)^{15-8} (-2 / x^{2})^{8}

= [6435 * 3^{8} * 2^{8 }* x^{7}] / x^{16}

= [6435 * 3^{7} * 2^{8}] / x^{9}

The two middle terms are [-6435 * 3^{8} * 2^{7}] / x^{6} and [6435 * 3^{7} * 2^{8}] / x^{9}.

**Question 5: Prove that if n is even, then in the expansion of (1 + x) ^{n}, coefficient of the middle term will be {[1. 3. 5…… (n – 1)] / [2. 4. 6 ……. n]} * 2^{n}. If n is odd, then the coefficient of both the middle term will be {[1. 3. 5…… (n)] / [2. 4. 6 ……. (n + 1)]} * 2^{n}.**

**Solution: **

If n is even in (1 + x) term we take (1 + x)^{2n} in place of (1 + x)^{n} then ‘n’ = 2n is also even.

Now, the middle term = [(2n / 2) + 1]^{th} term = (n + 1)^{th} term.

Similarly, if n is odd, then we take (1 + x)^{2n+1} in place of (1 + x)^{n} , then ‘n’ = 2n + 1 is also odd. Now middle term are

**Question 6: If in the expansion of (ax + (1 / bx)) ^{11}, the coefficient of x^{7} and x^{-7} are equal, then prove that ab – 1 – 0.**

**Solution: **

(r + 1)^{th} term in the expansion of (ax + (1 / bx))^{11} is given by

T_{r+1} = ^{11}C_{r} (ax)^{11-r} (1 / bx)^{r}

= ^{11}C_{r} [(a)^{11-r} / b^{r }] (x)^{11-r-r}

= ^{11}C_{r} [(a)^{11-r} / b^{r }] (x)^{11-2r}

In this term, the coefficient of x^{7} in the expansion of

^{11}C_{2} [(a)^{11-r} / b^{r }]

= 55 * [a^{9} / b^{2}] —– (1)

Similarly, for the coefficient of x^{-7},

-2r = -7 – 11

-2r = -18

r = 18 / 2

r = 9

The coefficient of x^{-7} in the expansion of

55 * [a^{9} / b^{2}] = 55 * [a^{2} / b^{9}]
[a^{9} / b^{9}] / [a^{2} / b^{2}] = 1

(ab)^{7} = 1

ab = 1

ab – 1 = 0

**Question 7: In the expansion of (1 + y) ^{n} if coefficients of 5^{th}, 6^{th} and 7^{th} terms are in A.P., then find the value of n.**

**Solution:**

**Question 8: In the binomial expansion of (x – a) ^{n}, the second, third and fourth terms are 250, 720 and 1080, respectively. Find x, a and n.**

**Solution:**

Second term in expansion of (x + a)^{n}

T_{2} = ^{n}C_{1} x^{n-1} a

According to the question,

^{n}C_{1} x^{n-1} a = 240 —- (1)

Similarly,

T_{3} = ^{n}C_{2} x^{n-2} a^{2} = 720 —- (2)

T_{4} = ^{n}C_{3} x^{n-3} a^{3} = 1080 —- (3)

On dividing (2) from (1), we get

^{n}C_{2} x^{n-2} a^{2} / ^{n}C_{1} x^{n-1} a = 720 / 240

On dividing (3) by (2), we get

[a / x] = 9 / 2(n – 2) —- (5)From (4) and (5), we get

6 / n – 1 = 9 / 2(n – 2)

n = 5

From 5x^{4}a = 240 and from (4)

a / x = 3 / 2

On solving both the equations, we get

x = 2, a = 3.

**Question 9: If in the expansion of (1 + a) ^{n} coefficient of three consecutive terms is in ratio 1: 7: 42, then find the value of n. **

**Solution:**

Let in the expansion of (1 + x)^{n} three consecutive terms are (r – 1 )^{th} term, r^{th} term and (r + 1)^{th} term.

Here, (r – 1)^{th} term = ^{n}C_{r }– 2a^{r-2} and its coefficient, = ^{n}C_{r} – 2

Similarly, coefficient of r^{th} term and (r + 1)^{th} term is ^{n}C_{r -1} and ^{n}C_{r} respectively.

Now, the given ratio of the coefficients is 1: 7: 42.

6r = n – r + 1

n – 7r + 1 = 0 …….(2)

On solving (1) and (2), we get n = 55.

### RBSE Maths Chapter 8: Exercise 8.3 Textbook Important Questions and Solutions

Question 1: If C_{0}, C_{1}, C_{2},… C_{n} are coefficients of expansion (1 + x)^{n} then find the value of :

(i) ^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8}

(ii) ^{8}C_{1}+ ^{8}C_{3} + ^{8}C_{5} + ^{8}C_{7}

Solution:

(i) ^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8}

We know that

^{n}C_{0} + ^{n}C_{1}+ ^{n}C_{2} + … + ^{n}C_{n} = 2^{n}

Putting n = 8

^{8}C_{0} + ^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8} = 2^{8}

⇒ 1 + ^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8} = 2^{8}

⇒ ^{8}C_{1} + ^{8}C_{2} + ^{8}C_{3} + … + ^{8}C_{8} = 2^{8} – 1 = 255.

(ii) ^{8}C_{1}+ ^{8}C_{3} + ^{8}C_{5} + ^{8}C_{7}

We know that ^{8}C_{1}+ ^{8}C_{3} + ^{8}C_{5} + ………. = 2^{n-1}

Putting n = 8

^{8}C_{1}+ ^{8}C_{3} + ^{8}C_{5} + ^{8}C_{7} = 2^{8 -1}

= 2^{7}

= 128.

**Question 2: If C _{0}, C_{1}, C_{2},… C_{n} are coefficients of expansion (1 + x)^{n }then prove that:**

**C _{0} + 3C_{1} + 5C_{2} + … + (2n + 1) C_{n} = (n + 1)2^{n}.**

**Solution:**

C_{0} + 3C_{1} + 5C_{2} + … + (2n + 1) C_{n} = (n + 1)2^{n}

= ^{n}C_{0} + (2 + 1) ^{n}C_{1} + (4 + 1) ^{n}C_{2} + … (2n + 1) ^{n}C_{n}

= ^{n}C_{0} + (2 * ^{n}C_{1} + ^{n}C_{1} + (4 * ^{n}C_{2} + ^{n}C_{2}) + … (2n * ^{n}C_{n} + ^{n}C_{n})

= 2 * ^{n}C_{1} + 4 * ^{n}C_{2} + 6 * ^{n}C_{3} + … + 2n * ^{n}C_{n} + (^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + ^{n}C_{3} + … ^{n}C_{n})

= 2 * [^{n}C_{1} + 2 ^{n}C_{2} + 3 * ^{n}C_{3} +… + ^{n}C_{n}] + (1 + 1)^{n }

= 2 [n + ((2n(n – 1) / [2!]) + (3 * [(n) (n – 1) (n – 2)] / 3!) + ……. n terms] + 2^{n}

= 2n [1 + (n – 1) + (n(n – 1) / [2!]) + ……. n terms] + 2^{n}

= 2n [1 + 1]^{n-1} + 2^{n}

= 2n * 2^{n-1} + 2^{n}

= n * 2^{n} + 2^{n}

= (n + 1) * 2^{n}

**Question 3: Prove that C _{0}C_{2} + C_{1}C_{3} + C_{2}C_{4} + … + C_{n-2} C_{n}= [2n!] / [n – 2]! [n + 2]!.**

**Solution:**

From Binomial theorem (1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + … + C_{r}x^{r} + … + C_{n}x^{n} … (i)

Again (x + 1)^{n} = C_{0}x^{n} + C_{1}x^{n-1} + C_{2}x^{n-2} + …+ C_{r}x^{n – r }+ …+ C_{n-1}x + C_{n} …(ii)

Multiplying equation (i) and (ii) we get

(1 + x)^{2n} = [C_{0} + C_{1}x + C_{2}x^{2} + … + C_{r}x^{r} + … + C_{n}x^{n} ] * [C_{0}x^{n} + C_{1}x^{n-1} + C_{2}x^{n-2} + …+ C_{r}x^{n – r }+ …+ C_{n-1}x + C_{n} …]

Comparing coefficient of x^{n – r} on both sides, we get

C_{0}C_{r} + C_{1}C_{r+1 }+ C_{2}C_{r +2} + … + C_{n-r}C_{n}

= ^{2n}C_{n-r} = [2n]! / [(n – r)! (2n – n + r)!]

= [2n]! / [(n – r)! (n + r)!] —– (iii)

Putting r = 2 in equation (iii),

C_{0}C_{2} + C_{1}C_{3} + C_{2}C_{4} + … + C_{n-2}C_{n} = ^{2n}C_{n-2} = [2n!] / [n – 2]! [2n – n + 2]!

= [2n!] / [n – 2]! [n + 2]!

**Question 4: C _{0} + 2C_{1} + 4C_{2} + 6C_{3} + … + 2nC_{n} = 1 + n2^{n}. **

**Solution:**

L.H.S.

= C_{0} + 2C_{1} + 4C_{2} + 6C_{3} + … + 2nC_{n}

= C_{0} + 2[C_{1} + 2C_{2} + 3C_{3} + … + nC_{n}]

= 1 + 2[n + [2n (n – 1) / 2!] + [3n (n – 1) (n – 2) / 3!] + ……. + n

= 1 + 2n [1 + (n + 1) + (n – 1) (n – 2) / 2! + ….. + 1]

= 1 + 2n (1 + 1)^{n-1}

= 1 + 2n * 2^{n-1}

= 1 + n * 2^{n-1+1}

= 1 + n * 2^{n}

**Question 5: [1 + (C _{1} / C_{0})] + [1 + C_{2} / C_{1}] + [1 + C_{3} / C_{2}] …….. + [1 + C_{n} / C_{n-1}] = [n + 1]^{n} / n!.**

**Solution:**

**Question 6: If expansion of (1 + x – 2x ^{2})^{6} is denoted by 1 + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + … + a_{12}x^{12} then prove that a^{2} + a^{4} + a^{6} + … + a^{12} = 31. **

**Solution: **

(1 + x – 2x^{2})^{6} = 1 + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + … + a_{12}x^{12} ……… (i)

Putting x = – 1 in equation (i)

{1 + 1 – 2(1)^{2}}^{6} = 1+ a_{1} + a_{2} + a_{3} + … + a_{12}

⇒ (2 – 2)^{6} = 1 + a_{1} + a_{2} + a_{3} + … + a_{12}

⇒ 1 + a_{1} + a_{2} + a_{3} + … + a_{12} = 0 … (ii)

Putting x = – 1 in equation (i)

{(1 – 1 – 2(-1)^{2}}^{6} = 1 – a_{1} + a_{2} + a_{3} + … + a_{12}

⇒ (-2)^{6} = 1 – a_{1} + a_{2} + a_{3} + … + a_{12}

⇒ 1 – a_{1} + a_{2} + a_{3} + … + a_{12} = 64 … (iii)

Adding equation (ii) and (iii) we get

⇒ 2 + 2a_{2} + 2a_{4} + … + a_{12} = 64

⇒ 2(1 + a_{2} + a_{4} + … + a_{12} = 64

⇒ 1 + a_{2} + a_{4} + … + a_{12} = 64 / 2 = 32

⇒ a_{2} + a_{4} + … + a_{12} = 32 – 1 = 31

Hence, a_{2} + a_{4} + … + a_{12} =31.

### RBSE Maths Chapter 8: Exercise 8.4 Textbook Important Questions and Solutions

**Question 1: Expand the binomial [1 + x ^{2}]^{-2} upto fourth term.**

**Solution:**

Expansion up to four terms of [1 + x^{2}]^{-2}

^{2}]

^{-2}= 1 + (-2) x

^{2}+ [(-2) (-2 – 1) (x

^{2})

^{2}/ [2!]] + [(-2) (-2 – 1) (-2 – 2) (x

^{2})

^{3}/ [3!]]

= 1 – 2x^{2} + [6x^{4} / 2!] + [(-2) (-3) (-4) x^{6}] / 3!

= 1 – 2x^{2} + [6x^{4} / 2!] – [24x^{6} / 3!]

= 1 – 2x^{2} + [6x^{4} / 2 * 1] – [24x^{6} / 3 * 2 * 1]

= 1 – 2x^{2} + 3x^{4} – 4x^{6}

**Question 2: Find the seventh term of the expansion (1 + x) ^{5/2}.**

**Solution:**

**Question 3: Find the general term of the expansion of (a ^{3} – x^{3})^{⅔}.**

**Solution:**

**Question 4: If x < 3, find the coefficient of (3 – x) ^{-8} in the expansion of (3 – x)^{-5}.**

**Solution: **

(x + 1)^{th} term in the expansion of (3 – x)^{-8}

**Question 5: Find the coefficient of x ^{6} in the expansion of (a + 2bx^{2})^{-3}. **

**Solution:**

(r + 1)^{th} term in the expansion of (a + 2bx^{2})^{-3}

For the coefficient of x^{6} in the this term 2r = 6 ⇒ r = 3

Hence, from equation (i), coefficient of x^{6} in the expansion of (a + 2bx^{2})^{-3}

= [(-1)^{3} 3 * 4 * 5 * 2^{3} * b^{3}] / [3! A^{3+3}]

= [-3 * 4 * 5 * 8 * b^{3}] / [3 * 2 * a^{6}]

= [-80b^{3} / a^{6}]

= -80 b^{3}a^{-6}

**Question 6: Find the coefficient of x ^{10} in the expansion of [1 + 3x^{2}] / [1 – x^{2}]^{3}.**

**Solution:**

**Question 7: Find the coefficient of x ^{r} in the expression of (1 – 2x + 3x^{2} – 4x^{3} + …)^{n} and if x = ½ and n = 1, then find the value of the expression. **

**Solution:**

We know that (1 + x)^{-2} = 1 – 2x + 3x^{2} – 4x^{3} +…+ (-1)^{r} (r + 1)x^{r} + ….. (i)

Given progression (1 – 2x + 3x^{2} – 4x^{3} + …)^{n}

= {(1 + x)^{-2}}^{n} [From equation (i)]

= (1 + x)-^{2n}

The (r + 1)^{th} term is

**Question 8: Prove that (1 + x + x ^{2} + x^{3} + …)^{2} = 1 + 2x + 3x^{2} + … **

**Solution:**

We know that 1 + x + x^{2} + x^{3} + … = (1 – x)^{-1} …. (i)

(1 + x + x^{2} + x^{3} + …)^{2} = {(1 – x)^{-1}}^{2} [From equation (i)] = (1 – x)^{2}

= (1 – x)^{2}

= 1 + 2x + [2(2 + 1)x^{2} / 2!] + …….

= 1 + 2x + 3x^{2} + ……

**Question 9: Prove that (1 + x + x ^{2} + x^{3} +….) (1 + 3x + 6x^{2} + …) = (1 + 2x + 3x^{2} + …)^{2} **

**Solution:**

We know that (1 – x)^{-1} = 1 + x + x^{2} + x^{3} +… … (i) and (1 – x)^{-3} = 1 + 3x + 6x^{2} + 10x^{3} + … (ii) L.H.S.

= (1 + x + x^{2} + x^{3} +…) (1 + 3x + 6x^{2} + …)

= {(1 – x)^{-1}} {(1 – x)^{-3}}

= {(1 – x)^{2}}^{-2}

= {(1 – x)^{-2}}^{2}

= (1 + 2x + 3x^{2} +…)^{2}

= R.H.S.

**Question 10: If x = 2y + 3y ^{2} + 4y^{3} + …then express y in a series of ascending powers of x. **

**Solution:**

⇒ x = 2y + 3y^{2} + 4y^{2} + …

⇒ x = (1 + 2y + 3y^{2} + 4y^{3} + …)^{-1} [∵ (1 – x)^{-2} = 1 + 2x + 3x^{2} + …]

⇒ (x + 1) = (1 – y)^{-2}

(1 – y^{2}) = 1 / (1 + x)

(1 – y^{2}) = (1 + x)^{-1}

Taking square root on both sides,

(1 – y) = (1 + x)^{-½ }

### RBSE Maths Chapter 8: Exercise 8.5 Textbook Important Questions and Solutions

**Question 1: If y is too small as compared to x, then prove that [x – y] ^{n} / [x + y]^{n} = 1 – [2ny / x], where y^{2} and higher power are negligible.**

**Solution:**

**Question 2: If x is too small such that x2 and higher powers are negligible then find the value of the expression {[9 + 2x] ^{½} (3 + 4x)} / [1 + x]^{⅕}.**

**Solution:**

{[9 + 2x]^{½} (3 + 4x)} / [1 + x]^{⅕}

= {[9]^{½} [1 + (2x / 9)]^{½} [3 + 4x]} / [1 + x]^{⅕}

= (3 [1 + ½ * [2x / 9] + {[(½) (½ – 1) * [(2 / 9) * x]^{2}} / 2! + ….. ] * [3 + 4x]) / [1 + (1 / 5) * x + ([1 / 5] – 1) (x^{2} / 2!)]

Ignoring x^{2} and terms of higher order,

= {3[1 + (x / 9)] [3 + 4x]} / [1 + (x / 5)]

= [5(9 + x) (3 + 4x)] / [3(5 + x)]

= {5(27 + 3x + 36x + 4x^{2})} / {3(5 + x)}

= 5(9 + 3x) / (5 + x)

= (45 + 65x) (5 + x)^{-1}

= 9 + 13x – (9x / 5) – (13x^{2} / 5)

= 9 + (56 / 5)x

**Question 3: Find the value of **

**[i] √30 up to 4 decimal places.**

**[ii] (1.03) ^{⅓} up to 4 decimal places **

**Solution:**

**[i] √30 up to 4 decimal places**

= 6 (1 – 0.08333 – 0.00347 – 0.00029 + …]

= 6 (0.9122)

= 5.47746

**(ii) (1.03) ^{⅓} up to 4 decimal places**

(1.03)^{⅓} = (1 + 0.03)^{⅓}

**Question 4: If x is approximately equal to 1, then prove that**

**[i] [mx ^{m} – nx^{n}] / [m – n] = x^{m+n}**

**[ii] [ax ^{b} – bx^{a}] / [x^{b} – x^{a}] = [1] / [1 – x]**

**Solution:**

**[i] [mx ^{m} – nx^{n}] / [m – n] = x^{m+n} **

**[ii] [ax ^{b} – bx^{a}] / [x^{b} – x^{a}] = [1] / [1 – x]**

### RBSE Maths Chapter 8: Exercise 8.6 Textbook Important Questions and Solutions

**Question 1: 1 + [2 / 3] * [1 / 2] + {[2 * 5] / [3 * 6]} * [1 / 2 ^{2}] + …….. **

**Solution:**

On comparing the given series with 1 + nx + [{n (n – 1) x^{2}} / 2!] + ……..

nx = [2 / 3] * [1 / 2]

nx = [1 / 3]

n^{2}x^{2} = 1 / 9

1 / [n^{2}x^{2}] = 9 —- (1)

^{2}} / 2!] = {[2 * 5] / [3 * 6]} * [1 / 2

^{2}] [n(n – 1) / 2!] * x

^{2}= 5 / [9 * 2

^{2}] —- (2)

Multiply equation (1) and (2),

[n – 1] / 2n = 5 / 44n – 4 = 10n

6n = -4

n = -2 / 3

If n = [-2 / 3], then [-2 / 3] * x = [1 / 3]

x = [-1 / 2]

Hence the sum of the given series,

= (1 + x)^{n}

= (1 – (1 / 2))^{-⅔ }

= (1 / 2)^{-⅔ }

= (2^{2})^{⅓}

= (4)^{⅓}

**Question 2: 1 + [1 / 3] * [1 / 4] + {[(1 * 4) / (3 * 6)] * [1 / 4 ^{2}]} + …… **

**Solution:**

On comparing the given series with the given series,

1 + nx + [n(n – 1) / 2!] * [x^{2}] + ……

nx = 1 / 12

[1 / n^{2}x

^{2}] = 144 —- (1) [n(n – 1) x

^{2}/ 2] = [1 / 72] —- (2)

Multiply equation (1) and (2)

[n(n – 1) x^{2}/ 2] * [1 / n

^{2}x

^{2}] = [1 / 72] * 144

(n – 1) / 2n = 2

n – 1 = 4n

n = [-1 / 3]

From equation (1),

[-1 / 3] * x = [1 / 12]x = [-1 / 4]

The sum of given series,

(1 + x)^{n} = (1 – (1 / 4))^{-⅓}

= (3 / 4)^{-⅓}

= (4 / 3)^{⅓}