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Given x₅ on set Z₅ = {0,1,2,3,4}

Now,

1 ×₅ 1 = remainder obtained by dividing 1 × 1 by 5

= 1

3 ×₅ 4 = remainder obtained by dividing 3 × 4 by 5

= 2

4 ×₅ 4 = remainder obtained by dividing 4 × 4 by 5

= 1

Therefore, the composition table is below

(i) ⁸C₁ + ⁸C₂ + ⁸C₃ + … + ⁸C₈ 

We know that

ⁿC₀ + ⁿC₁+ ⁿC₂ + … + ⁿC_n = 2ⁿ

Putting n = 8

⁸C₀ + ⁸C₁ + ⁸C₂ + …. + ⁸C₈ = 2⁸

⇒ 1 +⁸C₁ + ⁸C₂ + ⁸C₃ + … + ⁸C₈ = 2⁸

⇒ ⁸C₁ + ⁸C₂ + …. + ⁸C₈ = 2⁸ – 1 = 255.

(ii) ⁸C₁ + ⁸C₃ + ⁸C₅ + ⁸C₇

We know that

⁸C₁ + ⁸C₃ + ⁸C₅ +… = 2^n-1

Putting n = 8

⁸C₁ + ⁸C₃ + ⁸C₅ + ⁸C₇ = 2^8 -1 = 2⁷ = 128.

Practice the Important MCQ Questions for Class 11, which are given here. In this part, students can figure out how to determine the answer for the Binomial Theorem. Clear every one of the essentials and prepare altogether for the test-taking assistance from Class 11 Maths Binomial Theorem Objective Questions.

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Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The coefficient of y in the expansion of (y² + c/y)⁵ is

(a) 10c
(b) 10c²
(c) 10c³
(d) None of these

2. (1.1)¹⁰⁰⁰⁰ is _____ 1000

(a) greater than
(b) less than
(c) equal to
(d) None of these

3. The fourth term in the expansion (x – 2y)¹² is

(a) -1670 x⁹ × y³
(b) -7160 x⁹ × y³
(c) -1760 x⁹ × y³
(d) -1607 x⁹ × y³

4. If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is

(a) 2
(b) 1/2
(c) 3
(d) 4

5. The greatest coefficient in the expansion of (1 + x)¹⁰ is

(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

6. The coefficient of xn in the expansion of (1 – 2x + 3x² – 4x³ + ……..)-n is

(a) (2n)!/n!
(b) (2n)!/(n!)²
(c) (2n)!/{2×(n!)²}
(d) None of these

7. The coefficient of xn in the expansion (1 + x + x² + …..)^-n is

(a) 1
(b) (-1)ⁿ
(c) n
(d) n+1

8. In the expansion of (a + b)ⁿ, if n is odd then the number of middle term is/are

(a) 0
(b) 1
(c) 2
(d) More than 2

9. The number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is

(a) 4815
(b) 4851
(c) 8451
(d) 8415

10. if n is a positive ineger then 2³ⁿn – 7n – 1 is divisible by

(a) 7
(b) 9
(c) 49
(d) 81

11. The coefficient of the middle term in the expansion of (2+3x)⁴ is:

(a) 5!
(b) 6
(c) 216
(d) 8!

12. The value of (126)^1/3 up to three decimal places is

(a) 5.011
(b) 5.012
(c) 5.013
(d) 5.014

13. The coefficient of x³y⁴ in (2x+3y²)⁵ is

(a) 360
(b) 720
(c) 240
(d) 1080

14. The integral part of \((8+3\sqrt7)^n\) is

(a) an odd integer
(b) an even integer
(c) zero
(d) nothing can be said.

15. If the third term in the expansion of \([x+x^{log_{10}}\;x]^5\),is 10⁶ then x may be

(a) 1
(b) \(\sqrt{10}\)
(c) 10
(d) 10^-2/5

16. The number of terms in the expansion of (y^1/5+x^1/10)⁵⁵, in which powers of x and y are free from radical signs are

(a) six
(b) twelve
(c) seven
(d) five

17. If x = 99⁵⁰ +100⁵⁰ and y= (101)⁵⁰ then

(a) x = y
(b) x<y 
(c) x>y
(d) None of these

18. If A and B are coefficients of xⁿ in the expansions of (1+x)²ⁿ and (1+x)²ⁿ⁻¹ respectively, then A/B is equal to

(a) 4
(b) 2
(c) 9
(d) 6

19. In the expansion of (1 + x)⁵⁰, the sum of the coefficients of odd powers of x is 

(a) 2²⁶
(b) 2⁴⁹
(c) 2⁵⁰
(d) 2⁵¹

20. Find an approximate value of (0.99)5 using the first three terms of its binomial expansion.

(a) 0.591
(b) 0.951
(c) 0.195
(d) None

Answer:

1. Answer: (c) 10c³

Explanation: Given, binomial expression is (y² + c/y)⁵

Now, T_r+1 = 5Cr × (y²)^5-r × (c/y)r

= 5Cr × y^10-3r × Cr

Now, 10 – 3r = 1

⇒ 3r = 9

⇒ r = 3

So, the coefficient of y = 5C³ × c³ = 10c³

2. Answer: (a) greater than

Explanation: Given, (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰

¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁ × (0.1) + ¹⁰⁰⁰⁰C₂ ×(0.1)² + other +ve terms

= 1 + 10000×(0.1) + other +ve terms

= 1 + 1000 + other +ve terms > 1000

So, (1.1)10000 is greater than 1000

3. Answer: (c) -1760 x⁹ × y³

Explanation: 4th term in (x – 2y)¹² = T4

= T₃₊₁

= ¹²C₃ (x)^12-3 ×(-2y)³

= ¹²C₃ x⁹ ×(-8y³)

= {(12×11×10)/(3×2×1)} × x⁹ ×(-8y³)

= -(2×11×10×8) × x⁹ × y³

= -1760 x⁹ × y³

4. Answer: (b) 1/2

Explanation: (1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..

Now, {m(m – 1)/2}x² = (-1/8)x²

⇒ m(m – 1)/2 = -1/8

⇒ 4m² – 4m = -1

⇒ 4m² – 4m + 1 = 0

⇒ (2m – 1)² = 0

⇒ 2m – 1 = 0

⇒ m = 1/2

5. Answer: (b) 10!/(5!)²

Explanation: The coefficient of xr in the expansion of (1 + x)¹⁰ is ¹⁰C_r and ¹⁰C_r is maximum for

r = 10/ 2

= 5

Hence, the greatest coefficient = ¹⁰C₅

= 10!/(5!)²

6. Answer: (b) (2n)!/(n!)²

Explanation: (1 – 2x + 3x² – 4x³ + ……..)^-n = {(1 + x)^-2}^-n

= (1 + x)²ⁿ

So, the coefficient of xnC₃ = ²ⁿC_n = (2n)!/(n!)²

7. Answer: (b) (-1)ⁿ

Explanation: (1 + x + x² + …..)-n = (1 – x)^-n

Now, the coefficient of x = (-1)ⁿ × ⁿC_n

= (-1)ⁿ

8. Answer: (c) 2

Explanation: In the expansion of (a + b)ⁿ,

if n is odd then there are two middle terms which are {(n + 1)/2}th term and {(n+1)/2 + 1}th term.

9. Answer: (b) 4851

Explanation: Given, x + y + z = 100;

where x ≥ 1, y ≥ 1, z ≥ 1

Let u = x – 1, v = y – 1, w = z – 1

where u ≥ 0, v ≥ 0, w ≥ 0

Now, equation becomes

u + v + w = 97

So, the total number of solution = 97+3-1C_3-1

= ⁹⁹C₂

= (99 × 98)/2

= 4851

10. Answer: (c) 49

Explanation: 2³ⁿ – 7n – 1 

= 2^{3 × n} – 7n – 1

= 8ⁿ – 7n – 1

= (1 + 7)ⁿ – 7n – 1

= {ⁿC₀ + ⁿC₁ 7 + ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ} – 7n – 1

= {1 + 7n + nC₂ 7² + …….. + ⁿC_n 7ⁿ} – 7n – 1

= ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ

= 49(ⁿC₂ + …….. + ⁿC_n 7^n-2) 

which is divisible by 49

So, 2³ⁿ – 7n – 1 is divisible by 49

11. Answer: (c) 216

Explanation: If the exponent of the expression is n, then the total number of terms is n+1.

Hence, the total number of terms is 4+1 = 5.

Hence, the middle term is the 3rd term.

Therefore, T₃ = ⁴C₂.(2)².(3x)²

T₃ = (6).(4).(9x²)

T₃ = 216x².

Therefore, the coefficient of the middle term is 216.

12. Answer: (c) 5.013

Explanation: (126)^1/3 can also be written as the cube root of 126.

Hence, (126)^1/3 is approximately equal to 5.013.

Hence, option (c) 5.013 is the correct answer.

13. Answer: (b) 720

Explanation: Given: (2x+3y²)⁵

Therefore, the general form for the expression (2x+3y²)⁵ is T_r+1 = ⁵Cr. (2x)^r.(3y²)^5-r

Hence, T₃₊₁ = ⁵C₃ (2x)³.(3y²)^5-3

T₄ = ⁵C₃ (2x)³.(3y²)²

T₄ = ⁵C₃.8x³.9y⁴

On simplification, we get

T₄ = 720x³y⁴

Therefore, the coefficient of x³y⁴ in (2x+3y²)² is 720. 

14. Answer: (b) an even integer

Explanation: Let (8+\(3\sqrt{1}\))ⁿ = p+f, where p∈I and f is a proper fraction and let (8+\(3\sqrt 1\))ⁿ = f′, a proper  fraction [∵0<8−\(3\sqrt 7\)<1]

Since (8+\(3\sqrt 7\))ⁿ+(8−\(3\sqrt 7\))ⁿ = p+f+f′ is an even integer

∴p+1 is even

∴p is an odd integer

15. Answer: (c) 10

Explanation: Put log10x =y, the given expression becomes (x+xy)⁵.

T₃ = ⁵C₂.x³(xy)² = 10x^3+2y =10⁶

⇒(3+2y)log₁₀x = 5log₁₀ 10 = 5

⇒ (3+2y)y = 5

⇒ y=1,−5/2

⇒log₁₀ x = 1log₁₀ x =−5/2

16. Answer: (a) six

Explanation: Given expansion is (y^1/5+x^1/10)⁵⁵

The general term is

T_r+1= ⁵⁵Cr(y^1/5)^55−r. (x^1/10)r

T_r+1 would free from radical sign if powers of j and x are integers.

i.e \(\frac{55-r}{5}\) and r/10 are integer.

⇒ r is multiple of 10.

Hence, r = 0,10,20,30,40,50

It is an A.P.

Thus, 50 = 0+(k−1)10

50 =10k−10

k = 6

Thus, the six terms of the given expansion in which x andy are free from radical signs.

17. Answer: (b) x<y

Explanation: (101)⁵⁰ − (99)⁵⁰ =(100+1)⁵⁰ − (100−1)⁵⁰

=2[⁵⁰C₁(100)⁴⁹+⁵⁰C₃(100)⁴⁷+......+⁵⁰C₄₉(100)]

>2.⁵⁰C₁.(100)⁴⁹ =2×50(100)⁴⁹=(100)⁵⁰

⇒(101)⁵⁰ >(99)⁵⁰ +(100)⁵⁰

⇒ y>x

⇒ x<y.

18. Answer: (b) 2

Explanation: Given, A= Coefficient of xⁿ in (1+x)²ⁿ

and B= Coefficient of xⁿ in (1+x)²ⁿ⁻¹

∴ A = ²ⁿC_n and  B=²ⁿ⁻¹C_n

\(\therefore \frac{A}{B}=\frac{^{2n}C_n}{2n-1}C_n=\frac{2n}{n}\)

= 2. 

19. Answer: (b) 2⁴⁹

Explanation: C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + …..C_n = 2n 

Therefore C₁ + C₃ + C₅ + ….. C_n = 1/2 × 2n 

= 1/2 × 2⁵⁰ 

= 2⁴⁹

20. Answer: (b) 0.951

Explanation: (0.99)5 =(1−0.01)⁵

= 1 - ⁵C₁ x (0.01) + ⁵C₂ x (0.01)²........

= 1 - 0.05 + 10 x 0.0001......

= 1.001 - 0.05

= 0.951

Click here to practice MCQ Questions for Binomial Theorem Equations 11

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(i) Here a₁ = a₂ = 1 = a_{n - 1} + a_{n - 2} (n ≥ 3) 

Putting n = 3, a₃ = a₂ + a₁ = 1 + 1 = 2 

Putting n = 4, a₄ = a₃ + a₂ = 2 + 1 = 3 

Putting n = 5, a₅ = a₄ + a₂ = 3 + 2 = 5 

Putting n = 6, a₆ = a₅ + a₄ = 5 + 3 = 8 

∴ First six terms of the sequence are 1, 1, 2, 3, 5, 8

(ii) Here a₁ = 4 and a_{n + 1} = 2na_n 

Putting n = 1, a₂ = 2 × 1 × a₁ = 2 × 1 × 4 = 8 

Putting n = 2, a₃ = 2 × 2 × a₂ = 4 × 8 = 32 

Putting n = 3, a₄ = 8 × 192 = 1536 

Putting n = 4, a₅ = 2 × 4 × a₄ = 8 × 192 = 1536 

Putting n = 5, a₆ = 2 × 5 × a₅ = 10 × 1536 = 15360

(b) a ≥ g

AM ≥ GM 

∴ a ≥ g

Answer : (b) (n + 1) 2ⁿ 

C₀ + 3C₁ + 5C₂ + ..... (2n + 1) C_n 

= (C₀ + C₁ + C₂ + ..... + C_n) + (2C₁ + 4C₂ + ..... + 2n C_n) 

= 2ⁿ + 2 (C₁ + 2C₂ + 3.C₃ + ..... + n C_n) 

= 2ⁿ + 2 × n . 2^n–1 

 = 2ⁿ + n . 2ⁿ 

= 2ⁿ (n + 1).

Given :

(1-3x+10x²)ⁿ

Sum of coefficients = a

a = (1 - 3 + 10)ⁿ

=(2³)ⁿ =(2ⁿ)³ (1+x²)ⁿ 

Sum of coefficients = b 

b = (1+1)ⁿ 

= 2ⁿ 

Put value of b in a; we get : 

a = b³

Given :

(1-3x+3x²-x³)⁸

Highest power is (x³)⁸ = x²⁴

And lowest power is x⁰

So the expansion contains all the terms ranging from 0 to 24 

Therefore,

Total number of terms = 25

Answer :(c) ⁵²C₄

\(^{47}C_4 + \sum_{r=1}^5 \)\(^{52-r}C_3\) 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁷C₃ + ⁴⁷C₄ 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁸C₃ + ⁴⁸C₄  (\(\because\) ⁿC_r + ⁿC_r+1 = ⁿ⁺¹C_r+1) 

= ⁵¹C₃ + ⁵⁰C₃ + ⁴⁹C₃ + ⁴⁹C₄ 

= ⁵¹C₃ + ⁵⁰C₃ + ⁵⁰C₄ 

= ⁵¹C₃ + ⁵¹C₄ 

= ⁵²C₄

(102)⁶ = (100 + 2)⁶ 

= (100)⁶ + ⁶C₁(1 00)⁵.2 + ⁶C₂(100)⁴ .2²+ ⁶C₃(100)³ .2³ + ⁶C₄(100)² .2⁴ + ⁶C₅.100.2⁵ + ⁶C₆2⁶ 

= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64 

= 1,126,162,419,264

(1.0005)⁴ = (1 + 0.0005)⁴ 

= 14 + ⁴C₁(0.0005) + ⁴C₂(0.0005)² + ⁴C₃(0.0005)³ + ⁴C₄(0.0005)⁴ 1 + 0.002 + 0.0000015 + ……………. 

= 1.00200150 ≈ 1.0020

(0.99)⁴ = (1- 0.01)⁴ 

= ⁴C₀(0.01) – ⁴C₁ (0.01) + ⁴C₂(0.01)² – ⁴C₃(0.01)³ + ⁴C₄(0.01)⁴ 

= 1 – 0.04 + 0.0006 – 0.000004 + 0.00000001 

= 0.96059601 ≈ 0.9606

(i) (99)⁵

99 = (100- 1)

99⁵ = (100- 1)⁵ = [100+ (- 1)]⁵

By using Binomial theorem

99⁵ = ⁵C₀ (100)⁵ (-1)⁰ + ⁵C, (100)⁴ (-1)¹
+ ⁵C₂ (100)³ (- 1)² + ⁵C₃ (100)² (- 1)³
+ ⁵C₄ (100) (-1)⁴ + ⁵C₅ (100)⁰ (-1)⁵

= 1 x 10000000000 x 1 + 5 x 100000000
× (- 1) + 10 x 1000000 × 1 + 10 × 10000
× (- 1)+ 5 x 100 × 1 + 1 × 1 × (-1)

= 10000000000 – 500000000 +
10000000 – 100000 + 500 – 1

= 10010000500 – 500100001

= 9509900499

Hence, (99)⁵ = 9509900499

(ii) (1.1)⁶

(1.1)⁶ = (1 +0.1)⁶

= ⁶C₀(1)⁶ (0.1)⁰ + ⁶C₁ (1)⁵ (0.1)¹
+ ⁶C₂ (1)⁴ (0.1)² + ⁶C₃ (1)³ (0.3)³
+ ⁶C₄ (1)² (0.1)⁴ + ⁶C₅ (1)¹ (0.1)⁵ + ⁶C₆ (1)⁰ (0.1)⁶

= (1 x 1 x 1) + (6 x 1 x 0-1) + {15 x 1 x (0.1)²}
+ {20 x 1 x (0.1)³} + {15 x 1 x (0.1)⁴}
+ {6 x 1 x (0.1)⁵} + {1 x 1 x (0.1)⁶}

= 1 + 6 x 0.1 + 15 x (0.1)² + 20 x (0.1)³
+ 15 x (0.1)⁴+ 6 x (0.1)⁵ + (0.1)⁶

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

= 1+0.6 + 0.15 + 0020 + 0-0015 + 0.00006 + 0000001

= 1.771561

Answer : (d) 0

Given, (1 + x + x² + x³)ⁿ = \(\sum_{r=0}^{3n} a_r\, x^r\)

⇒ (1 + x + x² + x³)ⁿ = a₀ + a₁x + a₂x² + ..... + a_3n x³ⁿ

Putting x = –1 in the above-given equation, we have

a₀ – a₁ + a₂ – a₃ + ..... – a_3n = 0

(b) 2/3 

6/(1 - r) =  18 

6 = 18 - 18r

18r = 18 – 6 = 12 

r = 12/18 = 2/3

Given, 3⁴⁰⁰ 

= (1 + 2)⁴⁰⁰ 

=⁴⁰⁰ C₀ (1)^{400 – 0} (2)⁰ + ⁴⁰⁰ C₁ (1)^{400 – 1} (2)^{1 + 400}C₂ (1)^{400 – 2} (2)² +…+ ⁴⁰⁰ C₄₀₀ (1)^{400 – 400}(2)⁴⁰⁰ 

We observe that all term expect the first one are divisible by 100, so the remainder when 3 ⁴⁰⁰ + 100 is same as the [ ⁴⁰⁰ C₀ (1)⁴⁰⁰ (2)⁴⁰⁰  ] ÷ 100 = 1 ÷ 100

Answer : (a) - \(\frac{3}{10}\)

The Middle term in the expansion of (1 + αx)⁴ is the \((\frac{4}{2} +1)\)th term, i.e., 3rd term.

∴ t₃ = t _{2 + 1} = ⁴C₂ (αx)²

The Middle term in the expansion of (1 – αx)⁶ is the \((\frac{6}{2} +1)\)th term, i.e., 4th term. 

∴ T₄ = T_{3 + 1} = ⁶C₃ (– 1)³ (αx)³

∴ Coefficient of t ₃ = Coefficient of T₄

⇒ ⁴C₂ α² = ⁶C₃ (-1)³ α³

⇒ \(\frac{4\times 3}{2} \alpha^2 = (-1) \times \frac{6\times 5\times 4}{3\times 2} \alpha^3\)

⇒ 6α² = -20α³

⇒ α = - \(\frac{6}{20}\)  

= - \(\frac{3}{10}\) 

The general term of the expression is 

T_{r + 1} = ¹⁸C_r (3x² )^{18 – r} \((\frac{-1}{x})^{r}\) 

= ¹⁸C_r (-1)^r 3^18-r x ^36-3r

For coefficient of x¹²   36 – 3r = 12

 ∴ r = 8

∴ T₈ = ¹⁸C₈ (-1)⁸ 3^18-8 x^36-24

= ¹⁸C₈3¹⁰

9 term contains x¹² and coefficient is ¹⁸C₈3¹⁰

General term of \((2x^2-\frac{3}{x})^{11}\)

T_{r + 1} = ¹¹C_r (2x² )^{11 – r}  \((\frac{-3}{x})^{r}\)

= ¹¹C_r (-1)^r 2^11-r 3^r x^22-3r

For term contain x⁶ 

22 – 3r = 6

R = \(\frac{16}{3}\)

∴ r is not natural number

∴ Expansion of \((2x^2-\frac{3}{x})^{11}\) is not contain x⁶ term

Given as

(3x² – 1/x³)¹⁰

Lets consider the 7^th term as T₇

Therefore,

T₇ = T₆₊₁

= ¹⁰C₆ (3x²)^10-6 (-1/x³)⁶

= ¹⁰C₆ (3)⁴ (x)⁸ (1/x¹⁸)

= [10 × 9 × 8 × 7 × 81] / [4 × 3 × 2 × x¹⁰]

= 17010 / x¹⁰

∴ The 7^th term of the expression (3x² – 1/x³)¹⁰ is 17010 / x¹⁰.

(1 + x – 3x²)¹⁰ = 1 + a₁x + a₂x² + ..... + a₂₀x²⁰

Putting x = 1 in (i), we get:

1 + a₁ + a₂ + ..... + a₂₀ = (– 1)¹⁰ = 1

Putting x = – 1 in (i), we get

1 – a₁ + a₂ – ..... + a₂₀ = (– 3)¹⁰ = 3¹⁰

Adding (ii) and (iii), we get

2(1 + a₂ + a₄ + ..... + a₂₀) = 3¹⁰ + 1

⇒ 2(a₂ + a₄ + ..... + a₂₀) = 3¹⁰ – 1

⇒ a₂ + a₄ + ..... + a₂₀ = \(\frac{3^{10} -1}{2}\)

Correct option  (b)  153

Explanation:

Apply ^n+r–1C_r–1 to get number of terms ^{16 + 3 – 1}C_{3 – 1} = ¹⁸C₂

= 153

Correct option   (b) ³¹C₆ – ²¹C₆

Explanation:

Co-efficient of x⁵ in (1 + x)²¹ + (1 + x)²² +............+ (1 + x)³⁰ 

= Co-efficient of x⁵ in  (1 + x)²¹{(1 + x)¹⁰ -1}/(1 + x) - 1

coefficient of x⁶ in (1 + x)³¹ – (1 + x)²¹  is ³¹C₆ - ²¹C₆

Answer : (b) 0

¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇ 

= ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₉ – ¹⁵C₈          (∵ ⁿC_r = ⁿC_n–r) 

= 0

Answer : (d) 5 

(1 + x) ^–2 = 1 – 2x + 3x² – 4x³ + 5x⁴ 

⇒ Coefficient of x⁴ in the expansion of (1 + x)^–2 = 5.

Answer : (c) n+1

(1 + x + x² + x³ + .....)² 

= {(1 – x) ^–1)}² = (1 – x) ^–2 

= 1 + 2x + 3x² + 4x³ + ..... + (n + 1)xⁿ + .....

∴ Coefficient of xⁿ in this expansion 

=  (n + 1).

We have, 

Coefficient of \(x^n\) in \((1+x)(1-x)^n\)

= coefficient of \(x^n\) in \((1-x)^n\) + coefficient of \(x^{n-1}\) in \((1-x)^n\)

= \((-1)^n\) ⁿC_n+(-1)^n-1nC_n-1

=\((-1)^n(1-n)\)

Answer: (a) \(\frac{10!}{(5!)^2}\)  

(1 – 2x + 3x² – 4x³ + ..... ∞) ^–5 

= {(1 + x)^–2}^–5 = (1 + x)¹⁰ 

^∴ Coefficient of x5 = ¹⁰C₅ = \(\frac{10!}{5!\,5!}\) = \(\frac{10!}{(5!)^2}\) 

(∵ (1 + x)ⁿ = 1 + ⁿC₁ x + ⁿC₂ x² + ⁿC₃ x² + ....) 

Answer : (a) 1

(1 + 2x + 3x² + 4x³ + ....)^1/2  =  [(1 – x) ^–2]^1/2 

= (1 – x) ^–1  = 1 + x + x² + x³ + ..... 

⇒ coefficient of each term = 1 

⇒ coefficient of xⁿ = 1.

(i) False

(ii) True

(iii) False

(iv) False

(v) True

(vi) False

(vii) False

Given as

(3x – 1/x²)¹⁰

Now, the 5^th term from the end is the (11 – 5 + 1)th, is., 7^th term from the beginning.

Therefore,

T₇ = T₆₊₁

= ¹⁰C₆ (3x)^10-6 (-1/x²)⁶

= ¹⁰C₆ (3)⁴ (x)⁴ (1/x¹²)

= [10 × 9 × 8 × 7 × 81] / [4 × 3 × 2 × x⁸]

= 17010 / x⁸

Thus, the 5^th term of the expression (3x – 1/x²)¹⁰ is 17010 / x⁸.

Given as

(x^3/2 y^1/2 – x^1/2 y^3/2)¹⁰

Lets consider the 8^th term as T₈

Therefore,

T₈ = T₇₊₁

= ¹⁰C₇ (x^3/2 y^1/2)^10-7 (-x^1/2 y^3/2)⁷

= -[10×9×8]/[3×2] x^9/2 y^3/2 (x^7/2 y^21/2)

= -120 x⁸y¹²

∴ The 8^th term of the expression (x^3/2 y^1/2 – x^1/2 y^3/2)¹⁰ is -120 x⁸y¹².

Here x =7y² , a = - \(\frac{2}{y}\) and n = 12

T_r+1 = ¹²C_r (7Y²)^12-r .(\(\frac{-2}{y}\))^r

= ¹²C_r.7^12-r .y^24-2r. y^-r (-2)^r 

T_r+1 = ¹²C_r.7^12-r .(-2)^r .y^24-3r 

To find the coefficient of y³ equate the power of y or 3 

i.e., 24 – 3r = 3 ⇒ 21 = 3r ⇒ r= 7 

∴T₇₊₁ = ¹²C₇ .7^12-7 . (-2)⁷ y³ 

= ^-12C₇ .7⁵ 2⁷ . y3 

∴Coefficient of y³ is ^-12C₇.7⁵ .2⁷

Here, x → x² , a → \(\frac{-6}{x}\) and r = 15 

T_r+1 = ¹⁵C_r.(x²)^15-r(\(\frac{-6}{x}\))^r

T_r+1 = ¹⁵C_r.x^30-2r.(-6)^r .x^-r 

= ¹⁵C_r.(-6)^r .x^30-2r-r 

= ¹⁵C_r(-6)^r .x^30-3r . 

To find the coefficient of x¹⁸,equate the power of x to 18 

∴ 30 – 3r = 18 

⇒ 30 – 18 = 3r 

⇒ 3r = 12 

⇒ r = 4 

T₄₊₁ = ¹⁵C₄(-6) ⁴x¹⁸ 

∴ Coefficient of x¹⁸ is ¹⁵C₄. (6)⁴

\((3x - \frac{2}{x^2})^{15}\)

Here x → 3x a → \(\frac{-2}{x^2}\) and n =15 

T_r+1 = ¹⁵C_r.(3x)^15-r . (\(\frac{-2}{x^2}\))^r

= ¹⁵C_r.3^15-r .(-2)^r . x^15-r 

We have x^15-3r. = x⁰ ⇒ 15 = 3r ⇒ r = 5 

T = ¹⁵C₅.3^15-5 .(-2)⁵ = ^-15C₅ .3¹⁰ 2⁵ 

∴ The term independent of x is ^-15C₅.3¹⁰.2⁵

 \((x^2 - \frac{2}{x^3})^5\)

Here x → x² a → \(\frac{-2}{x^3}\) = 5_r(x²)^{5 – r} .\((\frac{-2}{x^3})^r\) = 5_r.x^10-2r.(-2)^r 

= ⁵C_r(-2)^r .x^10-5r 

We have 10 – 5r = 0 ⇒ r = 2 

T₂₊₁ = T₃ = ⁵C₂(-2)² .x⁰ = 4. \(\frac{5.4}{2.1}\) = 40 

∴ The term independent of x is 40

The general term is given by

T_{r + 1} = ¹⁰²⁴C_r \((5^{\frac{1}{2}})^{1024-r} ​​(7^{\frac{1}{8}})^{r}\)

= ¹⁰²⁴C_r 5 ⁵¹²  C_r5 ⁵¹² −\(\frac{r}{2}-7^\frac{r}{8}\)

={ ¹⁰²⁴C_r 5 ^512-r} x \(5^\frac{r}{2}\times7^\frac{r}{8}\)

= { ¹⁰²⁴C_r 5 ^512-r}  x \((5^4\times7)^\frac{r}{8}\)

Clearly, Tr + 1 will be an integer if \(​​\frac{r}{8}\)is an integer such that 0 ≤ r ≤ 1024 R is multiple of 8 satisfying 0 ≤ r ≤ 1024 r = 0, 8, 16, 24,…, 1024 r can assumes 129 values.

Hence, there are 129 integral terms in the expansion of \((5^{\frac{1}{2}} + 7^{\frac{1}{4}})^{1024}\)

Given expansion is (1 + x² )⁸

(1 + x² )⁸ = 1 + ⁸C₁ (1)^{8 – 1} (x² )¹ + ⁸C₂ (1)^{8 – 2} (x² ) ² + ⁸C₃ (1)8 – 3 (x²) 3 +⁸C₄ (1)^{8 – 4} (x² ) 4 + ⁸C₅ (1)^{8 – 5} (x² )⁵ + ⁸C₆ (1)⁸ – 6x² )⁶ +⁸C₇ (1)^{8 – 7} (x² )⁷ +⁸C⁸ (1)^{8 – 8} (x²)⁸ 

= 1 + 8x² + 28x⁴ + 56x⁶ + 70x⁸ + 56x⁰ + 28x² + 8x⁴ + x¹⁶

Coefficient of x⁶= 56 

Coefficient of x⁴ = 28 

∴ Different between the coefficients of x⁶ and x⁴

 = 56 – 28 = 28

Given, (1 – x)² (2 + x)⁵

= (1 + x² - 2) (1 + ⁵C₁ 2^{5– 1} x¹ + ⁵C₂ 2 ^5–2 x² + ⁵C₃ 2 ^5–3 x³ + ⁵C₄ 2^{5 – 4}  x⁴ +⁵C₅ 2^{5 – 5} x⁵) 

= (1 + x² - 2x) (1 + 5. 2⁴ .x + 10. 2³ x² + 10 . 2² x³ + 5 . 2 . x⁴ + x⁵ )

= (1 + x² – 2x) (1 + 80x + 80x² + 40x³ + 10x⁴ + x 5 ) 

∴ Coefficient of x⁴ = 80 – 80 + 10 = 10

Let a, b ∈ Z

a * b = 3a + 7b

b * a = 3b + 7a

Now, a * b ≠ b * a

Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2

= 3 + 14

= 17

2 * 1 = 3 × 2 + 7 × 1

= 6 + 7

= 13

So, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

Hence, * is not commutative on Z.

(i) Let us check the commutativity of *

Let a, b ∈ Q – {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

So,

a * b = b * a, ∀ a, b ∈ Q – {-1}

Let us prove that associativity of *

Let a, b, c ∈ Q – {-1}, then,

a * (b * c) = a * (b + c + b c)

= a + (b + c + b c) + a (b + c + b c)

= a + b + c + b c + a b + a c + a b c

(a * b) * c = (a + b + a b) * c

= a + b + a b + c + (a + b + a b) c

= a + b + a b + c + a c + b c + a b c

So,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1}

So, * is associative on Q – {-1}.

(ii) Let e be the identity element in I⁺ with respect to *

Such that, a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e (1 + a) = 0 and e (1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} [because a ! = -1]

Thus, 0 is the identity element in Q – {-1} with respect to *.

(iii) Let a and b ∈ Q – {-1} be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b + ab = 0 and b + a + ba = 0

b (1 + a) = – a Q – {-1}

b = -a/1 + a Q – {-1} [because a ! = -1]

So, -a/1 + a is the inverse of a ∈ Q – {-1}

(i) Let us prove commutativity of *

Let a, b ∈ N

a * b = 1

b * a = 1

So,

a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

Let us prove associativity of *

Let a, b, c ∈ N

Then a * (b * c) = a * (1)

= 1

(a * b) * c = (1) * c

= 1

So, a * (b * c) = (a * b) * c for all a, b, c ∈ N

Hence, * is associative on N.

(ii) Let us commutativity of *

Let a, b ∈ N

a * b = (a + b)/2

= (b + a)/2

= b * a

So, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

Let us prove associativity of *

Let a, b, c ∈ N

a * (b * c) = a * (b + c)/2

= [a + (b + c)]/2

= (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c

= [(a + b)/2 + c]/2

= (a + b + 2c)/4

Thus, a * (b * c) ≠ (a * b) * c

If a = 1, b = 2, c = 3

1 * (2 * 3) = 1 * (2 + 3)/2

= 1 * (5/2)

= [1 + (5/2)]/2

= 7/4

(1 * 2) * 3 = (1 + 2)/2 * 3

= 3/2 * 3

= [(3/2) + 3]/2

= 4/9

So, there exist a = 1, b = 2, c = 3 ∈ N such that a * (b * c) ≠ (a * b) * c

Hence, * is not associative on N.

Let a, b, c ∈ Z

a * (b * c) = a * (bc + 1)

= a(bc + 1) + 1

= abc + a + 1

(a * b) * c = (ab + 1) * c

= (ab + 1)c + 1

= abc + c + 1

So, a * (b * c) ≠ (a * b) * c

Hence, * is not associative on Z.

(i) Given as a * b = 1.c.m. (a, b)

2 * 4 = l.c.m. (2, 4)

= 4

3 * 5 = l.c.m. (3, 5) 

= 15

1 * 6 = l.c.m. (1, 6)

= 6

(ii) Let us prove commutativity of *

Let a, b ∈ N

a * b = l.c.m (a, b)

= l.c.m (b, a)

= b * a

So

a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

Now let us prove associativity of *

Let a, b, c ∈ N

a * (b * c ) = a * l.c.m. (b, c)

= l.c.m. (a, (b, c))

= l.c.m (a, b, c)

(a * b) * c = l.c.m. (a, b) * c

= l.c.m. ((a, b), c)

= l.c.m. (a, b, c)

So

(a * (b * c) = (a * b) * c, ∀ a, b , c ∈ N

Hence, * is associative on N.

Let e be the identity element in I⁺ with respect to *

Such that, a * e = a = e * a, ∀ a ∈ I⁺

a * e = a and e * a = a, ∀ a ∈ I⁺

a + e = a and e + a = a, ∀ a ∈ I⁺

e = 0, ∀ a ∈ I⁺

Hence, 0 is the identity element in I⁺ with respect to *.

Let e be the identity element in I⁺ with respect to *

Such that, a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e(1 + a) = 0 and e(1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} [because a ! = -1]

Hence, 0 is the identity element in Q – {-1} with respect to *.

As we know, the coefficients of (r + 1)^th term in (1 + x)ⁿ⁺¹ is ⁿ⁺¹C_r

Therefore, sum of the coefficients of the r^th and (r + 1)^th terms in (1 + x)ⁿ is

(1 + x)ⁿ = ⁿC_r-1 + ⁿC_r

= ⁿ⁺¹C_r [since, ⁿC_r+1 + ⁿC_r = ⁿ⁺¹C_r+1]

Thus proved.

(1 – x + x²)ⁿ = a₀ + a₁x + a₂x² + … + a_2nx²ⁿ

At x = 1,

(1 – 1 + 1²)ⁿ = a₀ + a₁(1) + a₂(1)² + … + a_2n(1)²ⁿ 

a₀ + a₁ + a₂ + … + a_2n = 1 …(1)

At x = -1, 

(1 – (-1) + (-1)²)ⁿ = a₀ + a₁(-1) + a²(-1)² + … + a_2n(-1)²ⁿ 

a₀ - a₁ + a₂ - … + a_2n = 3ⁿ …(2) 

On adding eq.1 and eq.2 

(a₀ + a₁ + a₂ + … + a_2n) + (a₀ - a₁ + a₂ - … + a_2n) = 1 + 3ⁿ 

2(a₀ + a₂ + a₄ + … + a²ⁿ) = 1 + 3ⁿ 

a₀ + a₂ + a₄ + … + a_2n = \(\frac{1+3^n}{2}\)

(i) We have,

(96)³

Lets express the given expression as two different entities and apply the binomial theorem.

(96)³ = (100 – 4)³

= ³C₀ (100)³ (4)⁰ – ³C₁ (100)² (4)¹ + ³C₂ (100)¹ (4)² – ³C₃ (100)⁰ (4)³

= 1000000 – 120000 + 4800 – 64

= 884736

(ii) We have,

(102)⁵

Lets express the given expression as two different entities and apply the binomial theorem.

(102)⁵ = (100 + 2)⁵

= ⁵C₀ (100)⁵ (2)⁰ + ⁵C₁ (100)⁴ (2)¹ + ⁵C₂ (100)³ (2)² + ⁵C₃ (100)² (2)³ + ⁵C₄ (100)¹ (2)⁴ + ⁵C₅ (100)⁰ (2)⁵

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

= 11040808032

(iii) We have,

(101)⁴

Lets express the given expression as two different entities and apply the binomial theorem.

(101)⁴ = (100 + 1)⁴

= ⁴C₀ (100)⁴ + ⁴C₁ (100)³ + ⁴C₂ (100)² + ⁴C₃ (100)¹ + ⁴C₄ (100)⁰

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

(iv) We have,

(98)⁵

Lets express the given expression as two different entities and apply the binomial theorem.

(98)⁵ = (100 – 2)⁵

= ⁵C₀ (100)⁵ (2)⁰ – ⁵C₁ (100)⁴ (2)¹ + ⁵C₂ (100)³ (2)² – ⁵C₃ (100)² (2)³ + ⁵C₄ (100)¹ (2)⁴ – ⁵C₅ (100)⁰ (2)⁵

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

= 9039207968