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

Question

UpStudy AI · Accepted Answer

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:**

$$
\sum_{k=1}^{n-1} (3k^{2} + k) = n^{2}(n - 1).
$$
The sum is $$ n^{2}(n - 1) $$.

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