JEE Main Previous year questions: Mathematical Induction and Binomial Theorem PDF Download

Q.1. The greatest positive integer k, for which 49^k + 1 is a factor of the sum 49¹²⁵ + 49¹²⁴ + ...+ 49² + 49 + 1, is (2020)
(1) 32
(2) 63
(3) 60
(4) 65
Ans. (2)
Solution. We have
1 + 49 + 49² + .... + 49¹²⁵

Hence, the greatest positive integer value of k is 63.

Q.2. If the sum of the coefficients of all even powers of x in the product (1 + x + x² + ... + x²ⁿ)(1 - x + x² - x³ + ... + x²ⁿ) is 61, then n is equal to ____.
Ans. 30
Solution. We have
(1 + x + x² + ... + x²ⁿ)(1 - x + x² - x³ + ... + x²ⁿ) = a₀ + a₁x + a₂x² + a₃x³ + ... + a_4nx⁴ⁿ
Substituting x = 1, we get
a₀ + a₁ + a₂ + ... + a_4n = 2n + 1 (1)
Substituting x = -1 here, we get
a₀ - a₁ + a₂ - a₃ + ... + a_4n = 2n + 1 (2)
From Eqs. (1) and (2), we get
a₀ + a₂ + a₄ + ... + a_4n = 2n + 1 ...(3)
Now, 2n + 1 = 61 ⇒ n = 30

Q.3. The coefficient of x⁷ in the expression (1 + x)¹⁰ + x(1 +x)⁹ + x²(1 +x)⁸ + ... + x¹⁰ is (2020)
(1) 210
(2) 330
(3) 120
(4) 420
Ans. (2)
Solution. We have
= (1 + x)¹⁰ + x(1 +x)⁹ + x²(1 +x)⁸ + ... + x¹⁰

Therefore, the coefficient of x⁷ in the expression is


Q.4. If α and β be the coefficients of x⁴ and x² respectively in the expansion of  then (2020)
(1) α + β = 60
(2) α + β = -30
(3) α - β = 60
(4) α - β = -132
Ans. (4)
Solution. We have
(x + a)ⁿ + (x - a)ⁿ = 2(T₁ + T₃ + T₅ + ...)
= 2[⁶C₀x⁶ + ⁶C₂x⁴(x² - 1) + ⁶C₄x²(x² - 1)² + ⁶C₆(x² - 1)³]

= 64x⁶ - 96x⁴ + 36x²
Hence, α - β = - 96 - 36 = -132

Q.5. The coefficient of x⁴ in the expansion of (1 + x + x²)¹⁰ is ____. (2020)
Ans. 615
Solution. The general term of the expansion of (1 + x + x²)¹⁰ is

For coefficient of x⁴, β + 2γ = 4. So,
...(1)
...(2)
...(3)
Hence, the coefficient of x⁴ in the expansion of (1 + x + x²)¹⁰ is 210 + 360 + 45 = 615.

Q.6. In the expansion of if l₁ is the least value of the term independent of x when and l₂ is the least value of the term independent of x when then the ratio l₂ : l₁ is equal to (2020)
(1) 1 : 8
(2) 1 : 16
(3) 8 : 1
(4) 16 : 1
Ans. (4)
Solution. The term independent from x in expression  is

If So, l₁ = ¹⁶C₈ . 2⁸ ...(1)
If  So,  ...(2)
Hence, 

Q.7. The coefficient of t⁴ in the expansion of  (2019)
(1) 14    
(2) 15
(3) 10    
(4) 12
Ans. (2)
Solution. Consider the expression


Hence, the coefficient of

= 15

Q.8. If  then k equals: (2019)
(1) 400    
(2) 50
(3) 200    
(4) 100
Ans. (4)
Solution. Consider the expression,





∴ k=100

Q.9. If the third term in the binomial expansion of equals 2560, then a possible value of x is: (2019)
(1) 1/4
(2) 4√2    
(3) 1/8
(4) 2√2
Ans. (1)
Solution. Third term of



Q.10. The positive value of λ for which the co-efficient of x² in the expression  is 720, is: (2019)
(1) 4    
(2) 2√2
(3) √5    
(4) 3
Ans. (1)
Solution. Since, coefficient of x² in the expression is a constant term, then Coefficient of x² in x²= co-efficient of constant term in


Then, for constant term,

Coefficient of x² in expression = ¹⁰C₂λ² = 720

⇒λ = 4
Hence, required value of λ is 4.

Q.11. If  then K is equal to: (2019)
(1) (25)²    
(2) 2²⁵ - 1
(3) 2²⁴    
(4) 2²⁵ 
Ans. (4)
Solution.

Then, by comparison, K = 2²⁵ 

Q.12. The value of r for which  is maximum, is: (2019)
(1) 15    
(2) 20
(3) 11    
(4) 10
Ans. (2)
Solution. Consider the expression
For maximum value of above expression r should be equal to 20.

Which is the maximum value of the expression,
So, r = 20.

Q.13. The sum of the real values of x for which the middle term in the binomial expansion of equals 5670 is: (2019)
(1) 0    
(2) 6
(3) 4    
(4) 8
Ans. (1)
Solution. Middle Term,  term in the binomial expansion of


⇒ x⁸ - 81 = 0
∴ sum of all values of x = sum of roots of equation (x⁸ - 81 = 0)


Q.14. Let S_n = 1 + q + q² +.... + qⁿ and 
where q is a real number and q ≠ 1. If

¹⁰¹C₁ + ¹⁰¹C₂·S₁ + .... + ¹⁰¹C₁₀₀·S₁₀₀ = αT₁₀₀, then α is equal to: (2019)
(1) 2⁹⁹    
(2) 202
(3) 200    
(4) 2¹⁰⁰ 
Ans. (4)
Solution.






Q.15. Let(x+10)⁵⁰ + (x-10)⁵⁰ = a_o + a₁x+a₂x² + .... + a₅₀x⁵⁰, for all x ∈ R; then  is equal to: (2019)
(1) 12.50    
(2) 12.00
(3) 12.25    
(4) 12.75
Ans. (3)
Solution. 

= 12.25

Q.16. A ratio of the 5^th term from the beginning to the 5^th term from the end in the binomial expansion of  is: (2019)
(1)
(2)
(3)
(4)
Ans. (3)
Solution.
 
5^th term from beginning

and 5^th term from end T_11-5+1

∴ T₅ : T₇ =



Q.17. The total number is irrational terms in the binomial expansion of  is:    (2019)
(1) 55    
(2) 49
(3) 48    
(4) 54
Ans. (4)
Solution. Let the general term of the expansion

Then, for getting rational terms, r should be multiple of L.C.M. of (5, 10)
Then, r can be 0, 10, 20, 30, 40, 50, 60.
Since, total number of terms = 61
Hence, total irrational terms = 61 - 7 = 54

Q.18. The sum of the co-efficients of all even degree terms in x in the expansion of  is equal to:    (2019)
(1) 29    
(2) 32    
(3) 26    
(4) 24
Ans. (4)
Solution.

Hence, the sum of coefficients of even powers of
x = 2[1 - 15 + 15 + 15 - 3- 1] = 24

Q.19. The sum of the series 2^.20C₀ + 5^.20C₁ + 8^.20C₂ + 11^.20C₃ + ... + 62^.20C₂₀ is equal to:    (2019)
(1) 2²⁶
(2) 2²⁵
(3) 2²³
(4) 2²⁴ 
Ans. (2)
Solution.


Q.20. If the fourth term in the binomial expansion of  is equal to 200, and x > 1, then the value of x is:    (2019)
(1) 100    
(2) 10    
(3) 10³    
(4) 10⁴ 
Ans. (2)
Solution. The fourth term is equal to 200.

Taking log₁₀ on both sides and putting log₁₀ x = t

According to the question x > 1, ∴ x = 10.

Q.21. If the fourth term in the Binomial expansion of  is 20 x 8⁷, then a value of x is:    (2019)
(1) 8³    
(2) 8²    
(3) 8    
(4) 8^-2 
Ans. (2)
Solution.

Now, take log₈ on both sides, we get


Q.22. If some three consecutive coefficients in the binomial expansion of (x+1)ⁿ in powers of x are in the ratio 2:15:70, then the average of these three coefficients is:    (2019)
(1) 964    
(2) 232    
(3) 227    
(4) 625
Ans. (2)
Solution.


Q.23. If the coefficients of x² and x³ are both zero, in the expansion of the expression (1 + ax + bx²) (1-3x)¹⁵ in powers of x, then the ordered pair (a, b) is equal to:    (2019)
(1) (28,861)    
(2) (-54,315)
(3) (28,315)    
(4) (-21,714)
Ans. (3)
Solution. Given expression is
Co-efficient of x² = 0
Q.24. The smallest natural number n, such that the coefficient of x in the expansion of     (2019)
(1) 38    
(2) 58    
(3) 23    
(4) 35
Ans. (1)
Solution.

To find coefficient of x, 2n - 5r = 1
Given ⁿC_r = ⁿC₂₃ ⇒ r = 23 or n - r = 23
∴ n = 58 or n = 38
The minimum value is n = 38

Q.25. The coefficient of x¹⁸ in the product (1+x)(1-x)¹⁰(1+x+x²)⁹ is:    (2019)
(1) 84    
(2) -126    
(3) -84   
(4) 126
Ans. (1)
Solution. Given expression,


Q.26. If ²⁰C₁ + (2²) ²⁰C₂ +(3²) ²⁰C₃+ ...... + (20²) ²⁰C₂₀= A(2^β), then the ordered pair (A, P) is equal to:    (2019)
(1) (420, 19)    
(2) (420, 18)
(3) (380,18)    
(4) (380, 19)
Ans. (2)
Solution.

=420 x 2¹⁸
Now, compare it with R.H.S., A = 420 and β = 18

Q.27. The term independent of x in the expansion of  is equal to:    (2019)
(1) - 72    
(2) 36    
(3) - 36    
(4) - 108
Ans. (4)
Solution. Given expression is,

Term independent of x,

= -72+ 36 = -36

Q.28. The sum of the coefficients of all odd degree terms in the expansion of     (2018)
(1) -1
(2) 0
(3) 1
(4) 2
Ans. (4)
Solution.

Sum of coefficient of odd powers = 2(1 - 10 + 10) = 2.

Q.29. If n is the degree of the polynomial and m is the coefficient of xⁿ  in it, then the ordered pair (n,m) is  (2018)
(1) (12, (20)⁴)
(2) (8, 5(10)⁴)
(3) (24, (10)⁸)
(4) (12, 8(10)⁴)
Ans. (4)
Solution. 

After rationalizing the polynomial, we get


So, the degree of the polynomial is 12,
No, thew coefficient of x¹² is


Q.30. The coefficient of x² in the expansion of the product (2 - x²) · ((1 + 2x + 3x²)⁶ + (1 - 4x²)⁶) is: (2018)
(1) 107
(2) 108
(3) 155
(4) 106
Ans. (4)
Solution. (a) Let a = ((1 + 2x + 3x²)⁶ + (1 - 4x²)⁶)
∴ Coefficient of x² in the expansion of the product
(2 - x²) ((1 + 2x + 3x²)⁶+ (1 - 4x²)⁶)
= 2 (Coefficient of x² in a) - 1 (Constant of expansion)
In the expansion of ((1 + 2x + 3x²)⁶ + (1 - 4x²)⁶).
Constant = 1 + 1 = 2
Coefficient of x² = [Coefficient of x² in (⁶C₀(1 + 2x)⁶(3x²)⁰)] + [Cofficient of x² in (⁶C₁(1 + 2x)⁵ (3x²)¹)] - [⁶C₁ (4x²)] = 60 + 6 x 3 - 24 = 54
∴ The coefficient of x² in (2 - x²)((1 + 2x + 3x²)⁶+ (1 - 4x²)⁶)
=2 x 54 - 1 (2)= 108 - 2 = 106

Q.31. The value of (²¹C₁ - ¹⁰C₁) + (²¹C₂ - ¹⁰C₂) + (²¹C₃ - ¹⁰C₃) + (²¹C₄ - ¹⁰C₄) + ... + (²¹C₁₀ - ¹⁰C₁₀) is: (2017)
(1) 2²⁰ – 2¹⁰
(2) 2²¹ – 2¹¹
(3) 2²¹ – 2¹⁰
(4) 2²⁰ – 2⁹ 
Ans. (1)
Solution. 


Q.32. If (27)⁹⁹⁹ is divided by 7, then the remainder is: (2017)
(1) 3
(2) 1
(3) 6
(4) 2
Ans. (3)
Solution. 

Simplify the Base:

• First, divide 27 by 7.
• 27 divided by 7 is 3 with a remainder of 6.
• This means 27 leaves the same remainder as 6 when divided by 7.
• So, instead of calculating 27⁹⁹⁹, you can work with 6⁹⁹⁹.

Identify the Pattern in Powers of 6:
Calculate the first few powers of 6 and find their remainders when divided by 7.

• 6¹ is 6. When divided by 7, the remainder is 6.
• 6² is 36. When divided by 7, the remainder is 1.
• 6³ is 216. When divided by 7, the remainder is 6.
• 6⁴ is 1296. When divided by 7, the remainder is 1.

Pattern Observed: The remainders alternate between 6 and 1 every power.

Determine the Position in the Pattern:

• The pattern repeats every 2 powers.
• Since 999 is an odd number, it corresponds to the first position in the pattern.
• Therefore, the remainder for 6⁹⁹⁹ divided by 7 is the same as the remainder for 6¹, which is 6.
• The remainder when 27⁹⁹⁹ is divided by 7 is 6.

Q.33. The coefficient of x^-5 in the binomial expansion of where x ≠ 0, 1 is:    (2017)
(1) -1
(2) 4
(3) 1
(4) -4
Ans. (3)
Solution.
Since a³+1 = (a+1)(a²-a+1)

Q.34. If the number of terms in the expansion of , x ≠ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is:    (2016)
(1) 64
(2) 2187
(3) 243
(4) 729
Ans. (4)
Solution. Total number of terms = ⁿ⁺²C₂ = 28
⇒(n+2)(n+1) = 56
⇒ n=6
Sum of coefficients =(1-2+4)ⁿ = 3⁶ = 729

Q.35. The value of is equal to:    (2016)
(1) 1085
(2) 560
(3) 680
(4) 1240
Ans. (3)
Solution. 


Q.36. For x ∈ R, x ≠ –1, if (1 + x)²⁰¹⁶ + x (1 + x)²⁰¹⁵ + x² (1 + x)²⁰¹⁴ + …….. + x²⁰¹⁶ =,then a₁₇ is equal to:    (2016)




Ans. (3)
Solution. 
= 

Q.37. then n satisfies the equation    (2016)
(1) n² + n – 110 = 0
(2) n² + 5n – 84 = 0
(3) n² + 3n – 108 = 0
(4) n² + 2n – 80 = 0
Ans. (3)
Solution.


Q.38. If the coefficients of x^–2 and x^–4 in the expansion of  are m and n respectively, then m/n is equal to:    (2016)
(1) 5/4
(2) 4/5
(3) 27
(4) 182
Ans. (4)
Solution.