Question

express the sum using summation notation. use the lower limit of summation given and k for the index of summation.
4a + (4a + d) + (4a + 2d) + (4a + 3d) + \cdots + (4a + nd)

4a + (4a + d) + (4a + 2d) + (4a + 3d) + \cdots + (4a + nd) = \sum\limits_{k=0}^{\square} \square

Explanation:

Step1: Identify the pattern

The terms are \(4a + kd\) where \(k\) starts at 0 and goes up to \(n\). Let's check: when \(k = 0\), term is \(4a+0d = 4a\); \(k = 1\), \(4a + 1d=4a + d\);... \(k = n\), \(4a+nd\).

Step2: Determine upper limit

The last term is \(4a + nd\), so \(k\) goes from 0 to \(n\). So the summation is \(\sum_{k = 0}^{n}(4a + kd)\).

Answer:

\(\sum_{k = 0}^{\boldsymbol{n}} (4a + kd)\) (Upper limit: \(n\), Summand: \(4a + kd\))