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

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.