The problem asks to evaluate the summation $\sum_{k=1}^{30} \left(6(3k) - 3\right)$ using linearity of summation. Let us solve it step-by-step.

Step 1: Expand the terms in the summand

The expression inside the summation is: $6(3k) - 3 = 18k - 3.$

So, the summation becomes: $\sum_{k=1}^{30} \left(18k - 3\right).$

Step 2: Apply the linearity of summation

By the linearity property, the summation of a sum can be split into two separate summations: $\sum_{k=1}^{30} \left(18k - 3\right) = \sum_{k=1}^{30} 18k - \sum_{k=1}^{30} 3.$

Step 3: Factor constants out of the summations

We can factor out constants from each summation: $\sum_{k=1}^{30} 18k = 18 \sum_{k=1}^{30} k, \quad \text{and} \quad \sum_{k=1}^{30} 3 = 3 \cdot 30.$

Step 4: Use the formula for $\sum_{k=1}^n k$

The summation $\sum_{k=1}^n k$ is given by the formula: $\sum_{k=1}^n k = \frac{n(n+1)}{2}.$ For $n = 30$, this becomes: $\sum_{k=1}^{30} k = \frac{30(30+1)}{2} = \frac{30 \cdot 31}{2} = 465.$

Step 5: Substitute the results back

Substituting $\sum_{k=1}^{30} k = 465$ and $\sum_{k=1}^{30} 3 = 3 \cdot 30 = 90$, we get: $\sum_{k=1}^{30} \left(18k - 3\right) = 18 \cdot 465 - 90.$

Step 6: Simplify the calculations

First, compute $18 \cdot 465$: $18 \cdot 465 = 8370.$ Now subtract $90$: $8370 - 90 = 8280.$

Final Answer:

$\sum_{k=1}^{30} \left(6(3k) - 3\right) = 8280.$

Would you like a deeper breakdown of any step or further explanation?

Here are some follow-up questions for practice:

1. What is the formula for the sum of the first $n$ natural numbers?
2. How does linearity of summation simplify solving sums with multiple terms?
3. What happens if you change the range of summation to $k=1$ to $k=50$ instead?
4. How would the answer change if the expression inside were $6(3k) + 3$?
5. Can this technique be extended to sums involving quadratic terms like $k^2$?

Tip: Always simplify the terms inside a summation before applying formulas or properties like linearity. This reduces errors and makes computation easier!