Page No 12.27:
Question 1:
1 + 2 + 3 + ... + n = i.e. the sum of the first n natural numbers is .
Answer:
Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +...+ n =
By the principle of mathematical induction, P(n) is true for all nN.
Page No 12.27:
Question 2:
12 + 22 + 32 + ... + n2 =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 3:
1 + 3 + 32 + ... + 3n−1 =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 4:
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 5:
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 6:
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 7:
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 8:
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 9:
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 10:
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 11:
2 + 5 + 8 + 11 + ... + (3n − 1) =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 12:
1.3 + 2.4 + 3.5 + ... + n. (n + 2) =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 13:
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 14:
1.2 + 2.3 + 3.4 + ... + n (n + 1) =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 15:
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 16:
12 + 32 + 52 + ... + (2n − 1)2 =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.27:
Question 17:
a + ar + ar2 + ... + arn−1 =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 18:
a + (a + d) + (a + 2d) + ... (a + (n − 1) d) =
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 19:
52n −1 is divisible by 24 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 20:
32n+7 is divisible by 8 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
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Question 21:
52n+2 −24n −25 is divisible by 576 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 22:
32n+2 −8n − 9 is divisible by 8 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 23:
(ab)n = anbn for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 24:
n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 25:
72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 26:
2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 27:
11n+2 + 122n+1 is divisible by 133 for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 28:
Given for n ≥ 2, where a > 0, A > 0.
Prove that
Answer:
Page No 12.28:
Question 29:
Prove that n3 7n + 3 is divisible by 3 for all n N.
Answer:
Hence, n3 7n + 3 is divisible by 3 for all n N.
Page No 12.28:
Question 30:
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 1 for all n N.
Answer:
Hence, 1 + 2 + 22 + ... + 2n = 2n+1 1 for all n N.
Page No 12.28:
Question 31:
7 + 77 + 777 + ... + 777
Answer:
Let P(n) be the given statement.
Now,
Now, P(m + 1) = 7 + 77 + 777 +....+ 777...(m + 1) digits...7
Page No 12.28:
Question 32:
is a positive integer for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 33:
is a positive integer for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 34:
for all n ∈ N and
Answer:
We need to prove for all n ∈ N and using mathematical induction.
For n = 1,
LHS =
and
Therefore, the given relation is true for n = 1.
Now, let the given relation be true for n = k.
We need to prove that the given relation is true for n = k + 1.
Now,
Let:
.
Now,
Thus, for all n ∈ N and .
Page No 12.28:
Question 35:
Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N?
Answer:
Page No 12.28:
Question 36:
for all n ∈ N
Answer:
Let P(n) be the given statement.
Thus, we have:
Now,
Thus, P(m + 1) is true.
Hence, by mathematical induction is true for all n ∈ N
Page No 12.28:
Question 37:
for all n ≥ 2, n ∈ N
Answer:
Let P(n) be the given statement.
Thus, we have:
Page No 12.28:
Question 38:
x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.
Answer:
Let P(n) be the given statement.
Now,
Page No 12.28:
Question 39:
Answer:
Let P(n) be the given statement.
Page No 12.29:
Question 40:
[NCERT EXEMPLAR]
Answer:
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Question 41:
[NCERT EXEMPLAR]
Answer:
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Question 42:
Answer:
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Question 43:
Answer:
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Question 44:
Show by the Principle of Mathematical induction that the sum Sn of then terms of the series is given by
[NCERT EXEMPLAR]
Answer:
Page No 12.29:
Question 45:
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n N. [NCERT EXEMPLAR]
Answer:
Hence, the number of subsets of a set containing n distinct elements is 2n, for all n N.
Page No 12.29:
Question 46:
[NCERT EXEMPLAR]
Answer:
Page No 12.29:
Question 47:
[NCERT EXEMPLAR]
Answer:
Disclaimer: It should be k instead n in the denominator of . The same has been corrected above.
Page No 12.29:
Question 48:
[NCERT EXEMPLAR]
Answer:
Page No 12.29:
Question 49:
[NCERT EXEMPLAR]
Answer:
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Question 50:
Answer:
Page No 12.30:
Question 1:
Make the correct alternative in following question:
If xn 1 is divisible by x , then the least positive integral value of is
(a) 1 (b) 2 (c) 3 (d) 4
Answer:
Hence, the correct alternative is option (a).
Page No 12.30:
Question 2:
Make the correct alternative in the following question:
(a) 19 (b) 17 (c) 23 (d) 25
Answer:
Hence, the correct alternative is option (b).
Page No 12.30:
Question 3:
Make the correct alternative in the following question:
If is divisible by 9 for all n N, then the least positive integral value of is
(a) 5 (b) 3 (c) 7 (d) 1
Answer:
Hence, the correct alternative is option (a).
Page No 12.30:
Question 4:
Make the correct alternative in the following question:
Let P(n): 2n < (1 × 2 × 3 × ... × n). Then the smallest positive integer for which P(n) is true
(a) 1 (b) 2 (c) 3 (d) 4
Answer:
As, 2n < (1 × 2 × 3 × ... × n) is possible only when n 4
So, the smallest positive integer for which P(n) is true, is 4.
Hence, the correct alternative is option (d).
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Question 5:
Make the correct alternative in the following question:
A student was asked to prove a statement P(n) by induction. He proved P(k +1) is true whenever P(k) is true for all k > 5 N and also P(5) is true. On the basis of this he could conclude that P(n) is true.
(a) for all n N (b) for all n > 5 (c) for all n 5 (d) for all n < 5
Answer:
As, P(5) is true and
P(k + 1) is true whenever P(k) is true for all k > 5 N.
By the definition of the priniciple of mathematical induction, we get
P(n) is true for all n 5.
Hence, the correct alternative is option (c).
Page No 12.30:
Question 6:
Make the correct alternative in the following question:
If P(n): 49n + 16n + is divisible by 64 for n N is true, then the least negative integral value of is
(a) 3 (b) 2 (c) 1 (d) 4
Answer:
Hence, the correct alternative is option (c).
Page No 12.30:
Question 1:
If P(n) : "2 × 42n + 1 + 33n + 1 is divisible by λ for all n ∈ N" is true, then the value of λ is _____________.
Answer:
If P(n) = 2 × 42n + 1 + 33n + 1 is divisible by λ ∀ n ∈ N is true
for n = 1,
for n = 2,
Since common factor of P(1) and P(2) is 11
Hence 2 × 42n+1 + 33n+1 is divisible by 11
i.e λ = 11
Page No 12.30:
Question 2:
If P(n) : 2n < n!, n ∈ N, then P(n) is true for all n ≥ _____________.
Answer:
Given P(n) : 2n < n! ; n ∈ N
for n = 1,
P(1) : 2' < 1!
i.e 2 < 1
Which is not true
for n = 2,
P(2) : 22 = 4 < 2!
i.e 4 < 2
Which is not true
for n = 3,
P(3) : 23 < 3!
i.e 8 < 1 × 2 × 3
i.e 8 < 6
Which is again not true.
for n = 4,
P(4) i.e 24 < 4!
i.e 16 < 24
i.e a true statement.
P(5) : 25 < 5!
i.e 32 < 1 × 2 × 3 × 4 × 5
i.e 32 < 120
Which is also true
∴ P(n) : 2n < n! is true for n > 3
P(n) : 2n < n! is true i.e for n ≥ 4
Page No 12.30:
Question 3:
If P(n) : 2n < n!, n ∈ N, then P(n) is true for all n > _____________.
Answer:
Given P(n) : 2n < n! ; n ∈ N,
for n = 1,
P(1) : 2' < 1!
i.e 2 < 1
Which is not true
for n = 2,
P(2) : 22 = 4 < 2!
i.e 4 < 2
Which is not true
for n = 3,
P(3) : 23 < 3!
i.e 8 < 1 × 2 × 3
i.e 8 < 6
Which is again not true.
for n = 4,
P(4) i.e 24 < 4!
i.e 16 < 24
i.e a true statement.
P(5) : 25 < 5!
i.e 32 < 1 × 2 × 3 × 4 × 5
i.e 32 < 120
Which is also true
∴ P(n) : 2n < n! is true for n > 3
P(n) : 2n < n! is true for n > 3
Page No 12.30:
Question 4:
For each n ∈ N, 102n – 1 + 1 is divisible by _____________.
Answer:
For each n ∈ N,
Let P(n) : 102n–1 + 1
for n = 1
L.H.S = 102(1)–1 + 1
= 101 + 1
= 10 + 1
= 11
i.e P(1) = 11
Assume P(n) is true
for n = 2,
Similarly, assume that P(k) is divisible by 11.
Page No 12.30:
Question 5:
If P(n) : n! > 2n – 1, n ∈ N, then P(n) is true for all n > _____________.
Answer:
P(n) : n! > 2n – 1; n ∈ N
for n = 1,
P(1) : 1! > 21–1
i.e 1 > 2° = 1
i.e 1 > 1
which is false a statement
for n = 2
P(2) : 2! > 22–1
i.e 2 > 21
i.e 2 > 2
which is again a false statement.
for n = 3
P(3) : 3! > 23–1
i.e 6 > 22 = 4
i.e 6 > 4 which is true
for n = 4
P(4) : 4! > 24–1
i.e 24 > 23 = 8 which is true
Hence, P(n) : n! > 2n – 1 is true
for n > 2
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Question 6:
If P(n) : n3 – n is divisible by 6, n ∈ N, then P(n) is true for all n ≥ _____________.
Answer:
P(n) : n3 – n is divisible by 6; n ∈ N
for n = 1,
P(1) : (1)3 – 1 = 0 which is divisible by 6
for n = 2,
P(2) : (2)3 – 2
= 8 – 2 = 6 which is divisible by 6
for n = 3,
P(3) : (3)3 – 3
= 27 – 3
= 24; which is divisible by 6
Hence, P(n) is true for n ≥ 2
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Question 7:
If P(n) : n2 < 2n, n ∈ N, then P(n) is true for all n ≥ _____________.
Answer:
P(n) : n2 < 2n, n ∈ N
for n = 1,
P(1) : 1 < 2 which is true statement
for n = 2,
P(2) : 22 < 22 which is not true/false statement
for n = 3,
P(3) : (3)2 < 23
i.e 9 < 8 which is a false statement
for n = 4,
P(4) : (4)2 < 24
i.e 16 < 4 × 4 = 16
which is a false statement
for n = 5,
P(5) : (5)2 < 25
i.e 25 < 32
which is a true statement
for n = 6,
P(6) : (36) < 26
i.e 36 < 4 × 4 × 4 = 64
Which is a true statement
for n = 7, P(7) : 49 < 27 = 128 which is again true statement
Hence, P(n) : n2 < 2n is true for n ≥ 5
Page No 12.30:
Question 8:
If then P(n) is true for all n ≥ _____________.
Answer:
for n = 1,
which is a false statement.
for n = 2,
which is true
for n = 3
which is again true
Hence, .
Page No 12.30:
Question 1:
State the first principle of mathematical induction.
Answer:
Let P(n) be a given statement involving the natural number n such that
(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural number). This step is known as the Basis step.
(ii) If the statement (called Induction hypothesis) is true for n = k (where k is a particular but arbitrary natural number), then the statement is also true for n = k + 1,
i.e, truth of P(k) implies the truth of P(k + 1). This step is known as the Induction (or Inductive) step.
Then P(n) is true for all natural numbers n.
Note: The first principle of mathematical induction states that if the basis step and the inductive step are proven, then P(n) is true for all natural numbers.
Page No 12.30:
Question 2:
Write the set of value of n for which the statement P(n): 2n < n! is true.
Answer:
As, n! > 2n when n > 3.
So, the set of value of n for which the statement P(n): 2n < n! is true = {n N: n > 3}.
Page No 12.30:
Question 3:
State the second principle of mathematical induction.
Answer:
Second principle of mathematical induction:
Let P(n) be a given statement involving the natural number n such that
(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural number). This step is known as the Basis step.
(ii) If the statement (called Induction hypothesis) is true for 1 n k (where k is a particular but arbitrary natural number), then the statement is also true for n = k + 1,
i.e, truth of P(k) implies the truth of P(k + 1). This step is known as the Induction (or Inductive) step.
Then P(n) is true for all natural numbers n.
Note: The second principle of mathematical induction is completely equivalent to the first principle of mathematical induction which states that if the basis step and the inductive step are proven, then P(n) is true for all natural numbers.
But the only difference is in the inductive hypothesis step that we assume not only that the statement holds for n = k but also that it is true for all 1 n k.
Also, the base can be other natural number as well apart 1 in both the principles.
Page No 12.30:
Question 4:
If P(n): is divisible by for all n N is true, then find the value of .
Answer:
Page No 12.3:
Question 1:
If P (n) is the statement "n(n + 1) is even", then what is P(3)?
Answer:
We have:
P(n): n(n + 1) is even.
Now,
P(3) = 3(3 + 1) = 12 (Even)
Therefore, P(3) is even.
Page No 12.3:
Question 2:
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
Answer:
We have:
Page No 12.3:
Question 3:
If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
Answer:
We have:
Page No 12.3:
Question 4:
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
Answer:
Page No 12.3:
Question 5:
Given an example of a statement P (n) such that it is true for all n ∈ N.
Answer:
Proved:
)
Page No 12.3:
Question 6:
If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
Answer:
Page No 12.3:
Question 7:
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.
Answer:
Let P(n) be the statement 3n < n!.
For n = 1,
3n = 3 × 1 = 3
n! = 1! = 1
Now, 3 > 1
So, P(1) is not true.
For n = 2,
3n = 3 × 2 = 6
n! = 2! = 2
Now, 6 > 2
So, P(2) is not true.
For n = 3,
3n = 3 × 3 = 9
n! = 3! = 6
Now, 9 > 6
So, P(3) is not true.
For n = 4,
3n = 3 × 4 = 12
n! = 4! = 24
Now, 12 < 24
So, P(4) is true.
For n = 5,
3n = 3 × 5 = 15
n! = 5! = 120
Now, 15 < 120
So, P(5) is true.
Similarly, it can be verified that 3n < n! for n = 6, 7, 8, ... .
Thus, the statement P(n) : 3n < n! is true for all n ≥ 4 but P(1), P(2) and P(3) are not true.
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