(e) Show that: (1) $$ \sum_{r=1}^{3}(3 r+4)=\sum_{r=1}^{3} 3 r+\sum_{r=1}^{3} 4 $$

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To demonstrate the equality:

$$
\sum_{r=1}^{3} (3r + 4) = \sum_{r=1}^{3} 3r + \sum_{r=1}^{3} 4
$$

we can approach it both **algebraically** and by **evaluating each side numerically**. This will confirm that both sides of the equation are indeed equal.

---

### **1. Algebraic Proof**

The summation operator has a fundamental property called **linearity**, which allows us to distribute the summation over addition and scalar multiplication. Specifically:

$$
\sum_{r=1}^{n} (a \cdot f(r) + b) = a \cdot \sum_{r=1}^{n} f(r) + \sum_{r=1}^{n} b
$$

Applying this property to our problem:

$$
\sum_{r=1}^{3} (3r + 4) = \sum_{r=1}^{3} 3r + \sum_{r=1}^{3} 4
$$

**Breaking it down:**

1. **Distribute the Summation:**
   
   $$
   \sum_{r=1}^{3} (3r + 4) = \sum_{r=1}^{3} 3r + \sum_{r=1}^{3} 4
   $$
   
2. **Factor Out Constants:**
   
   - For the first sum, factor out the constant 3:
     
     $$
     \sum_{r=1}^{3} 3r = 3 \cdot \sum_{r=1}^{3} r
     $$
   
   - The second sum is a constant being added, so:
     
     $$
     \sum_{r=1}^{3} 4 = 4 + 4 + 4 = 12
     $$
   
3. **Combine the Results:**
   
   $$
   3 \cdot \sum_{r=1}^{3} r + 12
   $$

This shows algebraically that the original summation can be split into the sum of two separate summations, each multiplied by their respective constants.

---

### **2. Numerical Verification**

Let's compute both sides of the equation explicitly to verify their equality.

**Compute the Left-Hand Side (LHS):**

$$
\sum_{r=1}^{3} (3r + 4) = (3 \cdot 1 + 4) + (3 \cdot 2 + 4) + (3 \cdot 3 + 4)
$$

Calculate each term:

- When $$ r = 1 $$: $$ 3(1) + 4 = 3 + 4 = 7 $$
- When $$ r = 2 $$: $$ 3(2) + 4 = 6 + 4 = 10 $$
- When $$ r = 3 $$: $$ 3(3) + 4 = 9 + 4 = 13 $$

Add them together:

$$
7 + 10 + 13 = 30
$$

**Compute the Right-Hand Side (RHS):**

$$
\sum_{r=1}^{3} 3r + \sum_{r=1}^{3} 4 = [3(1) + 3(2) + 3(3)] + [4 + 4 + 4]
$$

Calculate each sum:

- $$ \sum_{r=1}^{3} 3r = 3(1) + 3(2) + 3(3) = 3 + 6 + 9 = 18 $$
- $$ \sum_{r=1}^{3} 4 = 4 + 4 + 4 = 12 $$

Add them together:

$$
18 + 12 = 30
$$

---

### **Conclusion**

Both the algebraic proof and the numerical verification confirm that:

$$
\sum_{r=1}^{3} (3r + 4) = \sum_{r=1}^{3} 3r + \sum_{r=1}^{3} 4 = 30
$$

Therefore, the equality holds true.
$$ \sum_{r=1}^{3} (3r + 4) = \sum_{r=1}^{3} 3r + \sum_{r=1}^{3} 4 $$ This is true because the summation is linear, allowing the terms to be separated.

(e) Show that: (1) \( \sum_{r=1}^{3}(3 r+4)=\sum_{r=1}^{3} 3 r+\sum_{r=1}^{3} 4 \)

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