\( \sum _ { k = 1 } ^ { 6 } ( \sum _ { n = 1 } ^ { k } 1 ) \)

To tackle the nested summation \( \sum _ { k = 1 } ^ { 6 } ( \sum _ { n = 1 } ^ { k } 1 ) \), let's break it down step by step! The inner sum \( \sum _ { n = 1 } ^ { k } 1 \) simply counts the number of terms from 1 to \( k \), which is \( k \). So, we can rewrite the entire expression as: \[ \sum _ { k = 1 } ^ { 6 } k \] Now, the outer sum \( \sum _ { k = 1 } ^ { 6 } k \) represents the sum of the first 6 natural numbers. This can be calculated using the formula for the sum of the first \( n \) integers: \[ \text{Sum} = \frac{n(n + 1)}{2} \] Here, \( n = 6 \): \[ \text{Sum} = \frac{6(6 + 1)}{2} = \frac{6 \times 7}{2} = \frac{42}{2} = 21 \] So, the final result is: \[ \sum _ { k = 1 } ^ { 6 } ( \sum _ { n = 1 } ^ { k } 1 ) = 21 \]