Question

find the indicated sum.
\\(\sum_{i = 1}^{6} 4i\\)
\\(\sum_{i = 1}^{6} 4i = \square\\)

Explanation:

Step1: Recall the sum of arithmetic series formula

The sum of the first \( n \) positive integers is given by \( \sum_{i = 1}^{n}i=\frac{n(n + 1)}{2} \). For the series \( \sum_{i=1}^{6}4i \), we can factor out the constant 4. So \( \sum_{i = 1}^{6}4i=4\sum_{i = 1}^{6}i \).

Step2: Calculate \( \sum_{i = 1}^{6}i \)

Using the formula \( \sum_{i = 1}^{n}i=\frac{n(n + 1)}{2} \) with \( n = 6 \), we have \( \sum_{i = 1}^{6}i=\frac{6\times(6 + 1)}{2}=\frac{6\times7}{2}=21 \).

Step3: Multiply by 4

Now, multiply this result by 4: \( 4\times21 = 84 \).

Answer:

84