1. Prove that 1^2+ 2^2+ 3^2 + โฏ + ๐^2 = ๐(๐+1)(2๐+1)รท6, โ๐๐ โ โ by using the principle of
mathematical induction
First, we will prove the base case for n = 1:
When n = 1, the left side of the equation is 1^2 = 1.
Plugging n = 1 into the right side of the equation, we get:
1(1 + 1)(2(1) + 1)/6 = 1(2)(3)/6 = 6/6 = 1.
Since the left side and the right side both equal 1 when n = 1, the base case is true.
Now, we assume that the equation holds for some positive integer k, where k โฅ 1:
1^2 + 2^2 + 3^2 + ... + k^2 = k(k + 1)(2k + 1)/6.
We will prove that the equation also holds for k + 1:
1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = (k + 1)[(k + 1) + 1][2(k + 1) + 1]/6.
Starting with the left side:
1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = k(k + 1)(2k + 1)/6 + (k + 1)^2.
Multiplying both terms by 6:
6[k(k + 1)(2k + 1)/6 + (k + 1)^2] = k(k + 1)(2k + 1) + 6(k + 1)^2.
Expanding the multiplication:
k(k + 1)(2k + 1) + 6(k + 1)^2 = 2k^3 + 3k^2 + k + 6k^2 + 12k + 6.
Combining like terms:
2k^3 + 9k^2 + 13k + 6.
Factoring out a common factor of 2:
2(k^3 + 4.5k^2 + 6.5k + 3).
Factoring the cubic term:
2(k + 1)(k + 2)(k + 3).
Now, we have:
2(k + 1)(k + 2)(k + 3).
We can rewrite the right side of the equation as:
(k + 1)((k + 1) + 1)(2(k + 1) + 1)/6.
This is equal to:
(k + 1)(k + 2)(2k + 3)/6.
Therefore, we have shown that:
1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = (k + 1)(k + 2)(2k + 3)/6.
By the principle of mathematical induction, the statement is true for all positive integers n.
To prove the formula using mathematical induction, we need to follow these steps:
Step 1: Base Case
First, we need to verify if the formula holds true for the base case, which is when n = 1. We can calculate the sum of squares for n = 1:
1^2 = 1
On the right side of the formula, plugging in n = 1 gives:
1(1 + 1)(2(1) + 1) / 6 = 1
Since both sides of the equation equal 1, the base case holds true.
Step 2: Inductive Hypothesis
Assume that the formula holds true for some arbitrary positive integer k, which means we assume that:
1^2 + 2^2 + 3^2 + ... + k^2 = k(k + 1)(2k + 1) / 6
Step 3: Inductive Step
Now, we need to prove that if the formula holds true for k, it also holds true for k+1.
We will add (k + 1)^2 to both sides of the assumed formula:
1^2 + 2^2 + 3^2 + ... + k^2 + (k + 1)^2 = k(k + 1)(2k + 1) / 6 + (k + 1)^2
Simplifying the right side of the equation:
k(k + 1)(2k + 1) / 6 + (k + 1)^2
= (k + 1)[k(2k + 1) / 6 + (k + 1)]
= (k + 1)[2k^2 + k + 6k + 6] / 6
= (k + 1)(2k^2 + 7k + 6) / 6
= (k + 1)(k + 2)(2k + 3) / 6
Now, we need to show that the above expression is equivalent to:
(k + 1)((k + 1) + 1)(2(k + 1) + 1) / 6
Expanding this expression:
(k + 1)(k + 2)(2k + 3) / 6
= (k + 1)(k + 2)(2k + 3) / 6
Since the expanded expression is the same as the expression on the right side of the assumed formula, we have proven that if the formula holds true for k, it also holds true for k+1.
Step 4: Conclusion
By the principle of mathematical induction, we have shown that the formula:
1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1) / 6
holds true for all positive integers n.
To prove the given statement using mathematical induction, we need to follow two steps:
Step 1: Verify the base case.
Step 2: Assume the statement is true for n=k and prove it for n=k+1.
Step 1: Base case (n=1)
When n=1, we have 1^2 = 1. On the right-hand side of the equation, we have (1)(1+1)(2(1)+1)/6 = 1. Hence, the base case holds true.
Step 2: Inductive step
Assume the statement is true for n=k, i.e., 1^2 + 2^2 + 3^2 + ... + k^2 = k(k+1)(2k+1)/6.
Now, we need to prove the statement for n=k+1, i.e., 1^2 + 2^2 + 3^2 + ... + (k+1)^2 = (k+1)((k+1)+1)(2(k+1)+1)/6.
Using the assumption, we can rewrite the left-hand side of the equation:
1^2 + 2^2 + 3^2 + ... + k^2 + (k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2
= [(k+1)(k(2k+1) + 6(k+1))]/6
= [(k+1)(2k^2 + k + 6k + 6)]/6
= [(k+1)(2k^2 + 7k + 6)]/6
= [(k+1)(k+2)(2k+3)]/6.
Simplifying the right-hand side of the equation, we get:
(k+1)((k+1)+1)(2(k+1)+1)/6
= (k+1)(k+2)(2k+3)/6.
Hence, if the assumption is true for n=k, then it is true for n=k+1.
By successfully verifying the base case and proving the inductive step, we can conclude that the statement holds true for all natural numbers n by using the principle of mathematical induction.