https://file.aiquickdraw.com/mathbot/file/172657413682988kzzt54.jpg

Let's work through the problems displayed on the board. Here's the breakdown:

### 1. **Sigma Notation Expansion**:
The sigma notation problem is:
\[
\sum_{k=0}^{7} 3k^2
\]

This means summing the expression \(3k^2\) from \(k = 0\) to \(k = 7\). Let's write out the full sum:

\[
3(0^2) + 3(1^2) + 3(2^2) + 3(3^2) + 3(4^2) + 3(5^2) + 3(6^2) + 3(7^2)
\]

Simplifying each term:

\[
3(0) + 3(1) + 3(4) + 3(9) + 3(16) + 3(25) + 3(36) + 3(49)
\]

\[
= 0 + 3 + 12 + 27 + 48 + 75 + 108 + 147
\]

Adding all the terms together:

\[
0 + 3 + 12 + 27 + 48 + 75 + 108 + 147 = 420
\]

So, the full sum is \(420\).

---

### 2. **Mathematical Induction Problem**:
The second part asks to prove by mathematical induction that the sum of the first \(n\) even numbers is:
\[
2 + 4 + 6 + \dots + 2n = n(n+1)
\]

#### Step 1: Base Case (n = 1)
For \(n = 1\), the left-hand side is \(2\), and the right-hand side is:
\[
1(1+1) = 1 \times 2 = 2
\]
So, the base case holds.

#### Step 2: Inductive Hypothesis
Assume that the formula holds for some \(n = k\), i.e.,
\[
2 + 4 + 6 + \dots + 2k = k(k+1)
\]

#### Step 3: Inductive Step
We need to prove that the formula holds for \(n = k + 1\), i.e.,
\[
2 + 4 + 6 + \dots + 2k + 2(k+1) = (k+1)((k+1)+1)
\]

Start with the left-hand side:
\[
2 + 4 + 6 + \dots + 2k + 2(k+1)
\]
By the inductive hypothesis, the sum of the first \(k\) terms is \(k(k+1)\), so:
\[
k(k+1) + 2(k+1)
\]
Factor out \((k+1)\):
\[
(k+1)(k + 2)
\]
This matches the right-hand side of the formula for \(n = k + 1\), which is \((k+1)((k+1)+1)\).

Therefore, the formula holds for \(n = k + 1\), completing the inductive step.

#### Conclusion:
By mathematical induction, the formula \(2 + 4 + 6 + \dots + 2n = n(n+1)\) is true for all \(n \geq 1\).

---

Do you need any further explanation or details?

Here are five related questions to explore further:
1. How would you generalize this sigma notation to higher powers of \(k\)?
2. Can you use a different method to prove the sum of the first \(n\) even numbers?
3. What happens if you sum the first \(n\) odd numbers instead? What's the formula?
4. How would you approach sigma notation with factorial terms?
5. Could you explore the sum of squares for odd integers instead of even?

**Tip:** In mathematical induction, always ensure you check the base case first and carefully manage the algebraic steps in the inductive step.

Tuliskan notasi sigma di bawah ini dengan bentuk penjumlahan lengkap: \(\sum_{k=0}^{7} 3k^2\), dan buktikan dengan induksi matematika bahwa \(2 + 4 + 6 + \dots + 2n = n(n+1)\).

Learn how to expand a sigma notation sum and prove a series summation using mathematical induction. The problem involves expanding \(\sum_{k=0}^{7} 3k^2\) and proving that \(2 + 4 + 6 + \dots + 2n = n(n+1)\) using step-by-step induction. Suitable for students in Grades 10-12.

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