1. Question: Prove that the sum of the first n natural numbers is given by the formula (n(n+1))/2 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the first natural number is 1, which can be calculated using the formula as (1(1+1))/2 = 1. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the first k natural numbers is given by (k(k+1))/2.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the first (k+1) natural numbers is given by ((k+1)((k+1)+1))/2.
Using the assumption, the sum of the first k natural numbers is (k(k+1))/2. Adding (k+1) to both sides, we get:
(k(k+1))/2 + (k+1) = ((k+1)((k+1)+1))/2
Simplifying the left-hand side:
(k(k+1) + 2(k+1))/2 = ((k+1)((k+1)+1))/2
(k^2 + k + 2k + 2)/2 = ((k+1)((k+1)+1))/2
(k^2 + 3k + 2)/2 = ((k+1)(k+2))/2
(k+1)(k+2)/2 = ((k+1)(k+2))/2
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula (n(n+1))/2 holds true for all natural numbers.
2. Question: Prove that the sum of the squares of the first n natural numbers is given by the formula (n(n+1)(2n+1))/6 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the squares of the first natural number is 1, which can be calculated using the formula as (1(1+1)(2(1)+1))/6 = 1. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the squares of the first k natural numbers is given by (k(k+1)(2k+1))/6.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the squares of the first (k+1) natural numbers is given by ((k+1)((k+1)+1)(2(k+1)+1))/6.
Using the assumption, the sum of the squares of the first k natural numbers is (k(k+1)(2k+1))/6. Adding (k+1)^2 to both sides, we get:
(k(k+1)(2k+1))/6 + (k+1)^2 = ((k+1)((k+1)+1)(2(k+1)+1))/6
Simplifying the left-hand side:
(k(k+1)(2k+1) + 6(k+1)^2)/6 = ((k+1)((k+1)+1)(2(k+1)+1))/6
(k^2 + k + 2k^2 + k + 6k + 6)/6 = ((k+1)((k+1)+1)(2(k+1)+1))/6
(3k^2 + 8k + 6)/6 = ((k+1)(k+2)(2k+3))/6
(k+1)(k+2)(2k+3)/6 = ((k+1)(k+2)(2k+3))/6
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula (n(n+1)(2n+1))/6 holds true for all natural numbers.
3. Question: Prove that for any positive integer n, 3^n – 1 is divisible by 2 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the statement holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, 3^1 – 1 = 2, which is divisible by 2. Hence, the statement holds true for n = 1.
Assumption: Let’s assume that the statement holds true for some arbitrary value of k, i.e., 3^k – 1 is divisible by 2.
Proof: Now, we need to prove that the statement holds true for k+1, i.e., 3^(k+1) – 1 is divisible by 2.
Using the assumption, we can write 3^k – 1 as (2m), where m is an integer.
Now, let’s consider 3^(k+1) – 1:
3^(k+1) – 1 = 3^k * 3 – 1 = 3^k * 2 + 3^k – 1
Since 3^k * 2 is divisible by 2 (as 3^k is divisible by 2 according to our assumption), and 3^k – 1 is also divisible by 2, the sum of these two terms will also be divisible by 2.
Therefore, we have proved that if the statement holds true for k, then it also holds true for k+1. Hence, by the principle of mathematical induction, the statement 3^n – 1 is divisible by 2 holds true for all positive integers n.
4. Question: Prove that for any positive integer n, the sum of the cubes of the first n natural numbers is given by the formula ((n(n+1))/2)^2 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the cubes of the first natural number is 1, which can be calculated using the formula as ((1(1+1))/2)^2 = 1. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the cubes of the first k natural numbers is given by ((k(k+1))/2)^2.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the cubes of the first (k+1) natural numbers is given by (((k+1)((k+1)+1))/2)^2.
Using the assumption, the sum of the cubes of the first k natural numbers is ((k(k+1))/2)^2. Adding (k+1)^3 to both sides, we get:
((k(k+1))/2)^2 + (k+1)^3 = (((k+1)((k+1)+1))/2)^2
Simplifying the left-hand side:
((k(k+1))^2 + 4(k+1)^3)/4 = (((k+1)((k+1)+1))/2)^2
((k^2(k+1)^2 + 4(k+1)^3))/4 = (((k+1)((k+1)+1))/2)^2
(k^2(k+1)^2 + 4(k+1)^3)/4 = (((k+1)(k+2))/2)^2
(k+1)^2(k^2 + 4(k+1))/4 = (((k+1)(k+2))/2)^2
((k+1)^2(k^2 + 4k + 4))/4 = (((k+1)(k+2))/2)^2
(k+1)^2(k+2)^2/4 = (((k+1)(k+2))/2)^2
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula ((n(n+1))/2)^2 holds true for all positive integers n.
5. Question: Prove that for any positive integer n, the sum of the first n odd numbers is given by the formula n^2 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the first odd number is 1, which can be calculated using the formula as 1^2 = 1. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the first k odd numbers is given by k^2.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the first (k+1) odd numbers is given by (k+1)^2.
Using the assumption, the sum of the first k odd numbers is k^2. Adding (2k+1) to both sides, we get:
k^2 + (2k+1) = (k+1)^2
Simplifying the left-hand side:
k^2 + 2k + 1 = (k+1)^2
(k+1)^2 = (k+1)^2
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula n^2 holds true for all positive integers n.
6. Question: Prove that for any positive integer n, the sum of the reciprocals of the first n natural numbers is given by the formula n/n+1 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the reciprocals of the first natural number is 1, which can be calculated using the formula as 1/1+1 = 1/2. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the reciprocals of the first k natural numbers is given by k/k+1.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the reciprocals of the first (k+1) natural numbers is given by (k+1)/(k+1)+1.
Using the assumption, the sum of the reciprocals of the first k natural numbers is k/k+1. Adding 1/(k+1) to both sides, we get:
k/k+1 + 1/(k+1) = (k+1)/(k+1)+1
Simplifying the left-hand side:
(k + 1)/(k + 1) = (k + 1)/(k + 1)
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula n/n+1 holds true for all positive integers n.
7. Question: Prove that for any positive integer n, the sum of the first n even numbers is given by the formula n(n+1) using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the first even number is 2, which can be calculated using the formula as 1(1+1) = 2. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the first k even numbers is given by k(k+1).
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the first (k+1) even numbers is given by (k+1)(k+1+1).
Using the assumption, the sum of the first k even numbers is k(k+1). Adding (k+1)*2 to both sides, we get:
k(k+1) + (k+1)*2 = (k+1)(k+1+1)
Simplifying the left-hand side:
k^2 + k + 2k + 2 = (k+1)(k+2)
k^2 + 3k + 2 = k^2 + 3k + 2
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula n(n+1) holds true for all positive integers n.
8. Question: Prove that for any positive integer n, the sum of the first n Fibonacci numbers is given by the formula F(n+2) – 1 using mathematical induction, where F(n) represents the nth Fibonacci number.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base cases, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base cases: For n = 1, the sum of the first Fibonacci number is 1, which can be calculated using the formula as F(1+2) – 1 = F(3) – 1 = 2 – 1 = 1. Hence, the formula holds true for n = 1.
For n = 2, the sum of the first two Fibonacci numbers is 1 + 1 = 2, which can be calculated using the formula as F(2+2) – 1 = F(4) – 1 = 3 – 1 = 2. Hence, the formula holds true for n = 2.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the first k Fibonacci numbers is given by F(k+2) – 1.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the first (k+1) Fibonacci numbers is given by F((k+1)+2) – 1.
Using the assumption, the sum of the first k Fibonacci numbers is F(k+2) – 1. Adding F(k+1) to both sides, we get:
F(k+2) – 1 + F(k+1) = F((k+1)+2) – 1
Simplifying the left-hand side:
F(k+2) + F(k+1) – 1 = F(k+3) – 1
Since the Fibonacci sequence is defined as F(n) = F(n-1) + F(n-2), we can rewrite the left-hand side as:
F(k+3) – 1 = F(k+3) – 1
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula F(n+2) – 1 holds true for all positive integers n.
9. Question: Prove that for any positive integer n, the sum of the first n square numbers is given by the formula (n(n+1)(2n+1))/6 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the first square number is 1, which can be calculated using the formula as (1(1+1)(2(1)+1))/6 = 1. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the first k square numbers is given by (k(k+1)(2k+1))/6.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the first (k+1) square numbers is given by ((k+1)((k+1)+1)(2(k+1)+1))/6.
Using the assumption, the sum of the first k square numbers is (k(k+1)(2k+1))/6. Adding (k+1)^2 to both sides, we get:
(k(k+1)(2k+1))/6 + (k+1)^2 = ((k+1)((k+1)+1)(2(k+1)+1))/6
Simplifying the left-hand side:
(k(k+1)(2k+1) + 6(k+1)^2)/6 = ((k+1)((k+1)+1)(2(k+1)+1))/6
(k^2(k+1) + 2k(k+1) + 6(k+1)^2)/6 = ((k+1)((k+1)+1)(2(k+1)+1))/6
(k^2(k+1) + 2k(k+1) + 6(k+1)^2)/6 = ((k+1)(k+2)(2k+3))/6
(k+1)(k+2)(2k+3)/6 = ((k+1)(k+2)(2k+3))/6
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula (n(n+1)(2n+1))/6 holds true for all positive integers n.
10. Question: Prove that for any positive integer n, the sum of the first n natural numbers multiplied by their consecutive odd numbers is given by the formula n^3 using mathematical induction.
Answer: To prove this using mathematical induction, we need to show that the formula holds true for the base case, assume it holds true for some arbitrary value of n, and then prove that it holds true for n+1.
Base case: For n = 1, the sum of the first natural number multiplied by its consecutive odd number is 1 * 1 = 1, which can be calculated using the formula as 1^3 = 1. Hence, the formula holds true for n = 1.
Assumption: Let’s assume that the formula holds true for some arbitrary value of k, i.e., the sum of the first k natural numbers multiplied by their consecutive odd numbers is given by k^3.
Proof: Now, we need to prove that the formula holds true for k+1, i.e., the sum of the first (k+1) natural numbers multiplied by their consecutive odd numbers is given by (k+1)^3.
Using the assumption, the sum of the first k natural numbers multiplied by their consecutive odd numbers is k^3. Adding (k+1)(2k+1) to both sides, we get:
k^3 + (k+1)(2k+1) = (k+1)^3
Simplifying the left-hand side:
k^3 + 2k^2 + k + 2k^2 + 3k + 1 = (k+1)^3
k^3 + 4k^2 + 4k + 1 = (k+1)^3
(k+1)^3 = (k+1)^3
Hence, we have proved that if the formula holds true for k, then it also holds true for k+1. Therefore, by the principle of mathematical induction, the formula n^3 holds true for all positive integers n.