Solve 2a_{n}=n^2+n-(n-1)+n-1

$$2 a _ { n } = n ^ { 2 } + n - ( n - 1 ) + n - 1$$

Solve for a_n

$$a_{n}=\frac{n\left(n+1\right)}{2}$$

Steps for Solving Linear Equation

To find the opposite of n-1, find the opposite of each term.

$$2a_{n}=n^{2}+n-n+1+n-1$$

Combine n and -n to get 0.

$$2a_{n}=n^{2}+1+n-1$$

Subtract 1 from 1 to get 0.

$$2a_{n}=n^{2}+n$$

Divide both sides by 2.

$$\frac{2a_{n}}{2}=\frac{n\left(n+1\right)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a_{n}=\frac{n\left(n+1\right)}{2}$$

Solve for n

$$n=\frac{\sqrt{8a_{n}+1}-1}{2}$$

$$n=\frac{-\sqrt{8a_{n}+1}-1}{2}\text{, }a_{n}\geq -\frac{1}{8}$$