\( \sum _{k=1}^{n-1}(3k^{2}+k) \)

To evaluate the summation \[ \sum_{k=1}^{n-1} (3k^{2} + k), \] we can break it down into two separate sums and then simplify each: \[ \sum_{k=1}^{n-1} (3k^{2} + k) = 3\sum_{k=1}^{n-1} k^{2} + \sum_{k=1}^{n-1} k. \] **1. Sum of \( k^2 \):** The formula for the sum of squares of the first \( m \) natural numbers is: \[ \sum_{k=1}^{m} k^2 = \frac{m(m + 1)(2m + 1)}{6}. \] Here, \( m = n - 1 \), so: \[ \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n - 1)}{6}. \] **2. Sum of \( k \):** The formula for the sum of the first \( m \) natural numbers is: \[ \sum_{k=1}^{m} k = \frac{m(m + 1)}{2}. \] Again, with \( m = n - 1 \): \[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}. \] **3. Combine the sums:** Substituting back into the original expression: \[ 3\sum_{k=1}^{n-1} k^{2} + \sum_{k=1}^{n-1} k = 3 \cdot \frac{(n-1)n(2n - 1)}{6} + \frac{(n-1)n}{2}. \] Simplify the coefficients: \[ = \frac{(n-1)n(2n - 1)}{2} + \frac{(n-1)n}{2}. \] Factor out \( \frac{(n-1)n}{2} \): \[ = \frac{(n-1)n}{2} \left(2n - 1 + 1\right) = \frac{(n-1)n \cdot 2n}{2}. \] Simplify further: \[ = n^2 (n - 1). \] **Final Answer:** \[ \sum_{k=1}^{n-1} (3k^{2} + k) = n^{2}(n - 1). \]