Evaluate sum _(i=1)^n(i^2+1)= ) - (2n^3+3n^2+7n)/(6) (2n^3+3n^2+7)/(6) (3n^3+2n^2+7n)/(6) infty (5n^3+7n)/(6)

## Step 1: Expand the Summation<br />### The summation can be split into two separate sums: the sum of $i^2$ and the sum of 1. This allows us to utilize known summation formulas.<br />## Step 2: Apply Summation Formulas<br />### We know that $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^{n} 1 = n$. Substituting these formulas into the expanded summation yields $\frac{n(n+1)(2n+1)}{6} + n$.<br />## Step 3: Simplify the Expression<br />### Expanding and simplifying the expression $\frac{n(n+1)(2n+1)}{6} + n$ gives us $\frac{2n^3 + 3n^2 + n}{6} + \frac{6n}{6} = \frac{2n^3 + 3n^2 + 7n}{6}$.