MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers
Principle of Mathematical Induction Class 11 Maths MCQs Questions with Answers
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Class 11 Maths Chapter 4 Quiz
Class 11 Maths Chapter 4 MCQ Online Test
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Principle of Mathematical Induction Class 11 Maths MCQ online test
| Q1. | For all n∈N, 3n5 + 5n³ + 7n is divisible by |
A.5 |
|
B.15 |
|
C.10 |
|
D.3 |
Ans: 15
Given number = 3n5 + 5n² + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
Given number = 3n5 + 5n² + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
| Q2. | {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = |
A.1/(n + 1) for all n ∈ N. |
|
B.1/(n + 1) for all n ∈ R |
|
C.n/(n + 1) for all n ∈ N. |
|
D.n/(n + 1) for all n ∈ R |
Ans: 1/(n + 1) for all n ∈ N.
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
| Q3. | For all n ∈ N, 32n + 7 is divisible by |
A.8 |
|
B.3 |
|
C.11 |
|
D.non of these |
Ans: 8
Given number = 32n + 7
Let n = 1, 2, 3, 4, ……..
32n + 7 = 3² + 7 = 9 + 7 = 16
32n + 7 = 34 + 7 = 81 + 7 = 88
32n + 7 = 36 + 7 = 729 + 7 = 736
Since, all these numbers are divisible by 8 for n = 1, 2, 3, …..
So, the given number is divisible by 8
Given number = 32n + 7
Let n = 1, 2, 3, 4, ……..
32n + 7 = 3² + 7 = 9 + 7 = 16
32n + 7 = 34 + 7 = 81 + 7 = 88
32n + 7 = 36 + 7 = 729 + 7 = 736
Since, all these numbers are divisible by 8 for n = 1, 2, 3, …..
So, the given number is divisible by 8
| Q4. | The sum of the series 1 + 2 + 3 + 4 + 5 + ………..n is |
A.n(n + 1) |
|
B.(n + 1)/2 |
|
C.n/2 |
|
D.n(n + 1)/2 |
Ans: n(n + 1)/2
Given, series is series 1 + 2 + 3 + 4 + 5 + ………..n
Sum = n(n + 1)/2
Given, series is series 1 + 2 + 3 + 4 + 5 + ………..n
Sum = n(n + 1)/2
| Q5. | The sum of the series 1² + 2² + 3² + ……….. n² is |
A.n(n + 1) (2n + 1) |
|
B.n(n + 1) (2n + 1)/2 |
|
C.n(n + 1) (2n + 1)/3 |
|
D.n(n + 1) (2n + 1)/6 |
Ans: n(n + 1) (2n + 1)/6
Given, series is 1² + 2² + 3² + ……….. n²
Sum = n(n + 1)(2n + 1)/6
Given, series is 1² + 2² + 3² + ……….. n²
Sum = n(n + 1)(2n + 1)/6
| Q6. | For all positive integers n, the number n(n² − 1) is divisible by: |
A.36 |
|
B.24 |
|
C.6 |
|
D.16 |
Ans: 6
Given,
number = n(n² − 1)
Let n = 1, 2, 3, 4….
n(n² – 1) = 1(1 – 1) = 0
n(n² – 1) = 2(4 – 1) = 2 × 3 = 6
n(n² – 1) = 3(9 – 1) = 3 × 8 = 24
n(n² – 1) = 4(16 – 1) = 4 × 15 = 60
Since all these numbers are divisible by 6 for n = 1, 2, 3,……..
So, the given number is divisible 6
Given,
number = n(n² − 1)
Let n = 1, 2, 3, 4….
n(n² – 1) = 1(1 – 1) = 0
n(n² – 1) = 2(4 – 1) = 2 × 3 = 6
n(n² – 1) = 3(9 – 1) = 3 × 8 = 24
n(n² – 1) = 4(16 – 1) = 4 × 15 = 60
Since all these numbers are divisible by 6 for n = 1, 2, 3,……..
So, the given number is divisible 6
| Q7. | If n is an odd positive integer, then aⁿ + bⁿ is divisible by : |
A.a² + b² |
|
B.a + b |
|
C.a – b |
|
D.none of these |
Ans: a + b
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
| Q8. | n(n + 1) (n + 5) is a multiple of ____ for all n ∈ N |
A.2 |
|
B.3 |
|
C.5 |
|
D.7 |
Ans: 3
Let P(n): n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k): k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1) (k + 5) = 3m for some natural number m, …… (i)
Now, (k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)
= k(k + 1) (k + 2) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) + 3(k + 1) (k +4) [on simplification]
= 3m + 3(k + 1 ) (k + 4) [using (i)]
= 3[m + (k + 1) (k + 4)], which is a multiple of 3
⇒ P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Let P(n): n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k): k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1) (k + 5) = 3m for some natural number m, …… (i)
Now, (k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)
= k(k + 1) (k + 2) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) + 3(k + 1) (k +4) [on simplification]
= 3m + 3(k + 1 ) (k + 4) [using (i)]
= 3[m + (k + 1) (k + 4)], which is a multiple of 3
⇒ P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
| Q9. | For any natural number n, 7ⁿ – 2ⁿ is divisible by |
A.3 |
|
B.4 |
|
C.5 |
|
D.7 |
Ans: 5
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 72 – 22 = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 72 – 22 = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
| Q10. | The sum of the series 1³ + 2³ + 3³ + ………..n³ is |
A.{(n + 1)/2}² |
|
B.{n/2}² |
|
C.n(n + 1)/2 |
|
D.{n(n + 1)/2}² |
Ans: 4Q 11.12343Q 12.12343Q 13.12342Q 14.
| Q11. | (1² + 2² + …… + n²) _____ for all values of n ∈ N |
A.= n³/3 |
|
B.< n³/3 |
|
C.> n³/3 |
|
D.None of these |
Ans: > n³/3
Let P(n): (1² + 2² + ….. + n²) > n³/3.
When = 1, LHS = 1² = 1 and RHS = 1³/3 = 1/3.
Since 1 > 1/3, it follows that P(1) is true.
Let P(k) be true. Then,
P(k): (1² + 2² + ….. + k² ) > k³/3 …. (i)
Now,
1² + 2² + ….. + k²
+ (k + 1)²
= {1² + 2² + ….. + k² + (k + 1)²
> k³/3 + (k + 1)³ [using (i)]
= 1/3 ∙ (k³ + 3 + (k + 1)²) = 1/3 ∙ {k² + 3k² + 6k + 3}
= 1/3[k³ + 1 + 3k(k + 1) + (3k + 2)]
= 1/3 ∙ [(k + 1)³ + (3k + 2)]
> 1/3(k + 1)³
P(k + 1):
1² + 2² + ….. + k² + (k + 1)²
> 1/3 ∙ (k + 1)³
P(k + 1) is true, whenever P(k) is true.
Thus P(1) is true and P(k + 1) is true whenever p(k) is true.
Let P(n): (1² + 2² + ….. + n²) > n³/3.
When = 1, LHS = 1² = 1 and RHS = 1³/3 = 1/3.
Since 1 > 1/3, it follows that P(1) is true.
Let P(k) be true. Then,
P(k): (1² + 2² + ….. + k² ) > k³/3 …. (i)
Now,
1² + 2² + ….. + k²
+ (k + 1)²
= {1² + 2² + ….. + k² + (k + 1)²
> k³/3 + (k + 1)³ [using (i)]
= 1/3 ∙ (k³ + 3 + (k + 1)²) = 1/3 ∙ {k² + 3k² + 6k + 3}
= 1/3[k³ + 1 + 3k(k + 1) + (3k + 2)]
= 1/3 ∙ [(k + 1)³ + (3k + 2)]
> 1/3(k + 1)³
P(k + 1):
1² + 2² + ….. + k² + (k + 1)²
> 1/3 ∙ (k + 1)³
P(k + 1) is true, whenever P(k) is true.
Thus P(1) is true and P(k + 1) is true whenever p(k) is true.
| Q12. | {1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ……. + 1/{(2n + 1) (2n + 3)} = |
A.n/(2n + 3) |
|
B.n/{2(2n + 3)} |
|
C.n/{3(2n + 3)} |
|
D.n/{4(2n + 3)} |
Ans: n/{3(2n + 3)}
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 3}]
⇒ P(k + 1) is true, whenever P(k) is true.
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 3}]
⇒ P(k + 1) is true, whenever P(k) is true.
| Q13. | If n is an odd positive integer, then aⁿ + bⁿ is divisible by : |
A.a² + b² |
|
B.a + b |
|
C.a – b |
|
D.none of these |
Ans: a + b
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
| Q14. | (2 ∙ 7N + 3 ∙ 5N – 5) is divisible by ……….. for all N ∈ N |
A.6 |
|
B.12 |
|
C.18 |
|
D.24 |
Ans: 24
Let P(n): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
For n = 1, the given expression becomes (2 ∙ 71 + 3 ∙ 51 – 5) = 24, which is clearly divisible by 24.
So, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
⇒ (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) = 24m, for m = N
Now, (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5)
= (2 ∙ 7k ∙ 7 + 3 ∙ 5k ∙ 5 – 5)
= 7(2 ∙ 7k + 3 ∙ 5k – 5) = 6 ∙ 5k + 30
= (7 × 24m) – 6(5k – 5)
= (24 × 7m) – 6 × 4p, where (5k – 5) = 5(5k-1 – 1) = 4p
[Since (5k-1 – 1) is divisible by (5 – 1)]
= 24 × (7m – p)
= 24r, where r = (7m – p) ∈ N
⇒ P (k + 1): (2 ∙ 7k + 13 ∙ 5k + 1 – 5) is divisible by 24.
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Let P(n): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
For n = 1, the given expression becomes (2 ∙ 71 + 3 ∙ 51 – 5) = 24, which is clearly divisible by 24.
So, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
⇒ (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) = 24m, for m = N
Now, (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5)
= (2 ∙ 7k ∙ 7 + 3 ∙ 5k ∙ 5 – 5)
= 7(2 ∙ 7k + 3 ∙ 5k – 5) = 6 ∙ 5k + 30
= (7 × 24m) – 6(5k – 5)
= (24 × 7m) – 6 × 4p, where (5k – 5) = 5(5k-1 – 1) = 4p
[Since (5k-1 – 1) is divisible by (5 – 1)]
= 24 × (7m – p)
= 24r, where r = (7m – p) ∈ N
⇒ P (k + 1): (2 ∙ 7k + 13 ∙ 5k + 1 – 5) is divisible by 24.
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
| Q15. | For all n∈N, 52n − 1 is divisible by |
A.26 |
|
B.24 |
|
C.11 |
|
D.25 |
Ans: 24
Given number = 52n − 1
Let n = 1, 2, 3, 4, ……..
52n − 1 = 5² − 1 = 25 – 1 = 24
52n − 1 = 54 – 1 = 625 – 1 = 624 = 24 × 26
52n − 1 = 56 – 1 = 15625 – 1 = 15624 = 651 × 24
Since, all these numbers are divisible by 24 for n = 1, 2, 3, …..
So, the given number is divisible by 24
Given number = 52n − 1
Let n = 1, 2, 3, 4, ……..
52n − 1 = 5² − 1 = 25 – 1 = 24
52n − 1 = 54 – 1 = 625 – 1 = 624 = 24 × 26
52n − 1 = 56 – 1 = 15625 – 1 = 15624 = 651 × 24
Since, all these numbers are divisible by 24 for n = 1, 2, 3, …..
So, the given number is divisible by 24
| Q16. | 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = |
A.n(n + 1)(n + 2) |
|
B.{n(n + 1)(n + 2)}/2 |
|
C.{n(n + 1)(n + 2)}/3 |
|
D.{n(n + 1)(n + 2)}/4 |
Ans: {n(n + 1)(n + 2)}/3
Let the given statement be P(n). Then,
P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3){n(n + 1) (n + 2)}
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3){k(k + 1) (k + 2)}.
Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)
= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)
= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [using (i)]
= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)
= (1/3){(k + 1) (k + 2)(k + 3)}
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)
= (1/3){k + 1 )(k + 2) (k +3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N.
Let the given statement be P(n). Then,
P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3){n(n + 1) (n + 2)}
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3){k(k + 1) (k + 2)}.
Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)
= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)
= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [using (i)]
= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)
= (1/3){(k + 1) (k + 2)(k + 3)}
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)
= (1/3){k + 1 )(k + 2) (k +3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N.
| Q17. | 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} = |
A.{n(n + 3)}/{4(n + 1)(n + 2)} |
|
B.(n + 3)/{4(n + 1)(n + 2)} |
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C.n/{4(n + 1)(n + 2)} |
|
D.None of these |
Ans: {n(n + 3)}/{4(n + 1)(n + 2)}
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 2)}/{4(k + 2) (k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 2)}/{4(k + 2) (k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 2)}/{4(k + 2) (k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 2)}/{4(k + 2) (k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
| Q18. | For any natural number n, 7ⁿ – 2ⁿ is divisible by |
A.3 |
|
B.4 |
|
C.5 |
|
D.7 |
Ans: 5
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
| Q19. | The sum of n terms of the series 1² + 3² + 5² +……… is |
A.n(4n² – 1)/3 |
|
B.n²(2n² + 1)/6 |
|
C.none of these. |
|
D.n²(n² + 1)/3 |
Ans: n(4n² – 1)/3
Let S = 1² + 3² + 5² +………(2n – 1)²
⇒ S = {1² + 2² + 3² + 4² ………(2n – 1)² + (2n)²} – {2² + 4² + 6² +………+ (2n)²}
⇒ S = {2n × (2n + 1) × (4n + 1)}/6 – {4n × (n + 1) × (2n + 1)}/6
⇒ S = n(4n² – 1)/3
Let S = 1² + 3² + 5² +………(2n – 1)²
⇒ S = {1² + 2² + 3² + 4² ………(2n – 1)² + (2n)²} – {2² + 4² + 6² +………+ (2n)²}
⇒ S = {2n × (2n + 1) × (4n + 1)}/6 – {4n × (n + 1) × (2n + 1)}/6
⇒ S = n(4n² – 1)/3
| Q20. | For all n ∈ N, 3n5 + 5n³ + 7n is divisible by: |
A.5 |
|
B.15 |
|
C.10 |
|
D.3 |
Ans: 15
Given number = 3n5 + 5n³ + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
Given number = 3n5 + 5n³ + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
MCQ Questions for Class 11 Maths
-
MCQ Questions for Class 11 Maths Chapter 1 Sets
MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions
MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions
MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities
MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations
MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem
MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series
MCQ Questions for Class 11 Maths Chapter 10 Straight Lines
MCQ Questions for Class 11 Maths Chapter 11 Conic Sections
MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry
MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives
MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning
MCQ Questions for Class 11 Maths Chapter 15 Statistics
MCQ Questions for Class 11 Maths Chapter 16 Probability
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