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

Question

UpStudy AI · Accepted Answer

Certainly! Let's simplify the expression:

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

### Step 1: Break Down the Summation

First, separate the summation 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: Use Summation Formulas

We can use the standard formulas for the sum of the first $$ m $$ squares and the sum of the first $$ m $$ natural numbers:

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

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

Substitute $$ m = n - 1 $$ into the formulas:

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

### Step 3: Substitute Back into the Expression

Now, plug these back into the original expression:

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

Simplify each term:

$$
3 \times \frac{(n-1)n(2n - 1)}{6} = \frac{(n-1)n(2n - 1)}{2}
$$
$$
\frac{(n-1)n}{2} \text{ remains as is.}
$$

Combine the two terms:

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

Simplify further:

$$
\frac{(n-1)n(2n)}{2} = (n-1)n(n) = n^3 - n^2
$$

### Step 4: Add the Constant Term

Now, add the constant term $$ 4 $$ to the result:

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

### Final Simplified Expression

$$
4 + \sum_{k=1}^{n-1} (3k^2 + k) = n^3 - n^2 + 4
$$
The simplified expression is $$ n^3 - n^2 + 4 $$.

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