$$ 4+\sum _{k=1}^{n-1}(3k^{2}+k) $$

Question

UpStudy AI · Accepted Answer

Let's simplify the expression:

$$
4 + \sum_{k=1}^{n-1} (3k^2 + k)
$$

**Step 1: Break Down the Sum**

Separate the sum into two parts:

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

**Step 2: Apply Summation Formulas**

Use the standard formulas for the sums of the first $$ m $$ squares and the first $$ m $$ integers:

$$
\sum_{k=1}^{m} k^2 = \frac{m(m + 1)(2m + 1)}{6}
$$
$$
\sum_{k=1}^{m} k = \frac{m(m + 1)}{2}
$$

Here, $$ m = n - 1 $$:

$$
3 \sum_{k=1}^{n-1} k^2 = 3 \cdot \frac{(n-1)n(2n-1)}{6} = \frac{(n-1)n(2n-1)}{2}
$$
$$
\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}
$$

**Step 3: Combine the Results**

Add the two parts together:

$$
\frac{(n-1)n(2n-1)}{2} + \frac{(n-1)n}{2} = \frac{(n-1)n[(2n-1) + 1]}{2} = \frac{(n-1)n \cdot 2n}{2} = n^3 - n^2
$$

**Step 4: Add the Constant**

Finally, add the constant term:

$$
4 + (n^3 - n^2) = n^3 - n^2 + 4
$$

**Final Simplified Expression:**

$$
4 + \sum_{k=1}^{n-1} (3k^2 + k) = n^3 - n^2 + 4
$$

**Answer:** After simplifying, the expression equals n³ minus n² plus four. Thus,

n³ – n² + 4
The simplified expression is $$ n^3 - n^2 + 4 $$.

\( 4+\sum _{k=1}^{n-1}(3k^{2}+k) \)