Solve f(n)-f(n-1)=an+bn^2+cn^3

$$f ( n ) - f ( n - 1 ) = a n + b n ^ { 2 } + c n ^ { 3 }$$

Solve for a

$$\left\{\begin{matrix}a=-cn^{2}-bn+\frac{f}{n}\text{, }&n\neq 0\\a\in \mathrm{R}\text{, }&f=0\text{ and }n=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by n-1.

$$fn-\left(fn-f\right)=an+bn^{2}+cn^{3}$$

To find the opposite of fn-f, find the opposite of each term.

$$fn-fn+f=an+bn^{2}+cn^{3}$$

Combine fn and -fn to get 0.

$$f=an+bn^{2}+cn^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$an+bn^{2}+cn^{3}=f$$

Subtract bn^{2} from both sides.

$$an+cn^{3}=f-bn^{2}$$

Subtract cn^{3} from both sides.

$$an=f-bn^{2}-cn^{3}$$

Reorder the terms.

$$an=-cn^{3}-bn^{2}+f$$

The equation is in standard form.

$$na=f-bn^{2}-cn^{3}$$

Divide both sides by n.

$$\frac{na}{n}=\frac{f-bn^{2}-cn^{3}}{n}$$

Dividing by n undoes the multiplication by n.

$$a=\frac{f-bn^{2}-cn^{3}}{n}$$

Divide -cn^{3}-bn^{2}+f by n.

$$a=-cn^{2}-bn+\frac{f}{n}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{f-an-cn^{3}}{n^{2}}\text{, }&n\neq 0\\b\in \mathrm{R}\text{, }&f=0\text{ and }n=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by n-1.

$$fn-\left(fn-f\right)=an+bn^{2}+cn^{3}$$

To find the opposite of fn-f, find the opposite of each term.

$$fn-fn+f=an+bn^{2}+cn^{3}$$

Combine fn and -fn to get 0.

$$f=an+bn^{2}+cn^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$an+bn^{2}+cn^{3}=f$$

Subtract an from both sides.

$$bn^{2}+cn^{3}=f-an$$

Subtract cn^{3} from both sides.

$$bn^{2}=f-an-cn^{3}$$

Reorder the terms.

$$bn^{2}=-cn^{3}-an+f$$

The equation is in standard form.

$$n^{2}b=f-an-cn^{3}$$

Divide both sides by n^{2}.

$$\frac{n^{2}b}{n^{2}}=\frac{f-an-cn^{3}}{n^{2}}$$

Dividing by n^{2} undoes the multiplication by n^{2}.

$$b=\frac{f-an-cn^{3}}{n^{2}}$$

Divide -cn^{3}-an+f by n^{2}.

$$b=-cn+\frac{f-an}{n^{2}}$$