Question
How to establish an identity by induction related to a summation series?
Answers
GPT-4o mini
Establishing an identity by induction related to a summation series typically involves two main steps: the base case and the inductive step. Here’s how you can go through these steps systematically:
### Step 1: State the Identity
First, you need to formulate the identity or summation you wish to prove. For example, let’s say we want to prove the following formula for the sum of the first \( n \) positive integers:
\[
S(n) = \frac{n(n+1)}{2}
\]
### Step 2: Base Case
Verify the identity for a base case, usually for \( n = 1 \).
For our example:
\[
S(1) = \frac{1(1+1)}{2} = \frac{1 \cdot 2}{2} = 1
\]
This matches the direct calculation:
\[
1 = 1
\]
So the base case holds.
### Step 3: Inductive Hypothesis
Assume that the identity holds for some arbitrary positive integer \( k \). This is called the inductive hypothesis. For our example, we assume:
\[
S(k) = \frac{k(k+1)}{2}
\]
### Step 4: Inductive Step
Now, you want to prove that the identity also holds for \( k + 1 \). Start with the assumption \( S(k) \) and determine \( S(k+1) \):
\[
S(k+1) = S(k) + (k + 1)
\]
Substituting the inductive hypothesis into the equation gives:
\[
S(k+1) = \frac{k(k+1)}{2} + (k + 1)
\]
### Step 5: Simplify
Now, simplify the expression for \( S(k+1) \):
\[
S(k+1) = \frac{k(k+1)}{2} + \frac{2(k + 1)}{2}
\]
Combine the terms:
\[
S(k+1) = \frac{k(k+1) + 2(k + 1)}{2}
= \frac{(k + 1)(k + 2)}{2}
\]
### Step 6: Conclusion
Finally, notice that this result is exactly the right-hand side of the identity we wanted to prove for \( n = k + 1 \):
\[
S(k+1) = \frac{(k + 1)((k + 1) + 1)}{2}
\]
Since we've proven that the assumption holds for \( k + 1 \) given that it holds for \( k \) and we've verified the base case, we can conclude by the principle of mathematical induction that the identity
\[
S(n) = \frac{n(n+1)}{2}
\]
holds for all positive integers \( n \).
### Recap
1. **State the identity.**
2. **Verify the base case.**
3. **Assume it holds for \( n = k \) (inductive hypothesis).**
4. **Show it holds for \( n = k + 1 \).**
5. **Conclude that the identity holds for all positive integers.**
Using these steps, you can establish various summation identities or similar mathematical statements through mathematical induction.
### Step 1: State the Identity
First, you need to formulate the identity or summation you wish to prove. For example, let’s say we want to prove the following formula for the sum of the first \( n \) positive integers:
\[
S(n) = \frac{n(n+1)}{2}
\]
### Step 2: Base Case
Verify the identity for a base case, usually for \( n = 1 \).
For our example:
\[
S(1) = \frac{1(1+1)}{2} = \frac{1 \cdot 2}{2} = 1
\]
This matches the direct calculation:
\[
1 = 1
\]
So the base case holds.
### Step 3: Inductive Hypothesis
Assume that the identity holds for some arbitrary positive integer \( k \). This is called the inductive hypothesis. For our example, we assume:
\[
S(k) = \frac{k(k+1)}{2}
\]
### Step 4: Inductive Step
Now, you want to prove that the identity also holds for \( k + 1 \). Start with the assumption \( S(k) \) and determine \( S(k+1) \):
\[
S(k+1) = S(k) + (k + 1)
\]
Substituting the inductive hypothesis into the equation gives:
\[
S(k+1) = \frac{k(k+1)}{2} + (k + 1)
\]
### Step 5: Simplify
Now, simplify the expression for \( S(k+1) \):
\[
S(k+1) = \frac{k(k+1)}{2} + \frac{2(k + 1)}{2}
\]
Combine the terms:
\[
S(k+1) = \frac{k(k+1) + 2(k + 1)}{2}
= \frac{(k + 1)(k + 2)}{2}
\]
### Step 6: Conclusion
Finally, notice that this result is exactly the right-hand side of the identity we wanted to prove for \( n = k + 1 \):
\[
S(k+1) = \frac{(k + 1)((k + 1) + 1)}{2}
\]
Since we've proven that the assumption holds for \( k + 1 \) given that it holds for \( k \) and we've verified the base case, we can conclude by the principle of mathematical induction that the identity
\[
S(n) = \frac{n(n+1)}{2}
\]
holds for all positive integers \( n \).
### Recap
1. **State the identity.**
2. **Verify the base case.**
3. **Assume it holds for \( n = k \) (inductive hypothesis).**
4. **Show it holds for \( n = k + 1 \).**
5. **Conclude that the identity holds for all positive integers.**
Using these steps, you can establish various summation identities or similar mathematical statements through mathematical induction.