- Prove the following by using the principle of mathematical induction for all n € N:
1 + 2 + 3 + … + n < 1/8(2n+1)2
Solution:
Let the given statement be P(n), i.e.,
P(n): 1 + 2 + 3 + … + n < 1/8(2n+1)2
It can be noted that P(n) is true for n = 1 since .
1<1/8 (2.1 + 1)2 = 9/8
Let P(k) be true for some positive integer k, i.e.,
1 + 2 + 3 + … + k < 1/8(2k+1)2 ————(1)
We shall now prove that P(k+1) is true whenever P(k) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.
19: Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3.
Solution:
Let the given statement be P(n), i.e.,
P(n): n (n + 1) (n + 5), which is a multiple of 3.
It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.
Let P(k) be true for some positive integer k, i.e., k (k + 1) (k + 5) is a multiple of 3.
∴ k (k + 1) (k + 5) = 3m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.
- Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11.
Solution:
Let the given statement be P(n), i.e.,
P(n): 102n – 1 + 1 is divisible by 11.
It can be observed that P(n) is true for n = 1
since P(1) = 102.1 – 1 + 1 = 11, which is divisible by 11.
Let P(k) be true for some positive integer k, i.e., 102k – 1 + 1 is divisible by 11.
∴ 102k – 1 + 1 = 11m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Thus, P(k+1) is true whenever P(k) is true.
Hence by the principal of mathematical induction, statement P(n) is true for all natural numbers i.e., N.
- Prove the following by using the principle of mathematical induction for all n ∈ N:
x2n – y2n is divisible by x + y.
Solution:
Let the given statement be P(n), i.e.,
P(n): x2n – y2n is divisible by x + y.
It can be observed that P(n) is true for n = 1.
This is so because x2 × 1 – y2 × 1 = x2 – y2 = (x + y) (x – y) is divisible by (x + y).
Let P(k) be true for some positive integer k, i.e.,
x2k – y2k is divisible by x + y.
∴ Let x2k – y2k = m (x + y), where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider,
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.
22: Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8.
Solution:
Let the given statement be P(n), i.e.,
P(n): 32n + 2 – 8n – 9 is divisible by 8.
It can be observed that P(n) is true for n = 1
since 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8.
Let P(k) be true for some positive integer k, i.e., 32k + 2 – 8k – 9 is divisible by 8.
∴ 32k + 2 – 8k – 9 = 8m; where m ∈ N … (1) We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.
- Prove the following by using the principle of mathematical induction for all n ∈ N:
41n – 14n is a multiple of 27.
Solution:
Let the given statement be P(n), i.e.,
P(n): 41n – 14n is a multiple of 27.
It can be observed that P(n) is true for n = 1
since 41l – 14l = 27, which is a multiple of 27.
Let P(k) be true for some positive integer k, i.e.,
41k – 14k is a multiple of 27
∴ 41k – 14k = 27m, where m ∈ N ………………………………. (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
- Prove the following by using the principle of mathematical induction for all n€ N:
(2n +7) < (n + 3)2
Solution:
Let the given statement be P(n), i.e.,
P(n): (2n +7) < (n + 3)2
It can be observed that P(n) is true for n = 1
since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true.
Let P(k) be true for some positive integer k, i.e.,
(2k + 7) < (k + 3)2 … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.