Solve 1/4n^2+2/3+3-[1/4(n-1)^2+2/3(n-1)+3]

$$\frac { 1 } { 4 } n ^ { 2 } + \frac { 2 } { 3 } + 3 - [ \frac { 1 } { 4 } ( n - 1 ) ^ { 2 } + \frac { 2 } { 3 } ( n - 1 ) + 3 ]$$

Evaluate

$$-\frac{n}{6}+\frac{13}{12}$$

Solution Steps

Add \frac{2}{3} and 3 to get \frac{11}{3}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n-1\right)^{2}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n^{2}-2n+1\right)+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{1}{4} by n^{2}-2n+1.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{2}{3} by n-1.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}n-\frac{2}{3}+3\right)$$

Combine -\frac{1}{2}n and \frac{2}{3}n to get \frac{1}{6}n.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{1}{4}-\frac{2}{3}+3\right)$$

Subtract \frac{2}{3} from \frac{1}{4} to get -\frac{5}{12}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n-\frac{5}{12}+3\right)$$

Add -\frac{5}{12} and 3 to get \frac{31}{12}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}\right)$$

To find the opposite of \frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}, find the opposite of each term.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\frac{1}{4}n^{2}-\frac{1}{6}n-\frac{31}{12}$$

Combine \frac{1}{4}n^{2} and -\frac{1}{4}n^{2} to get 0.

$$\frac{11}{3}-\frac{1}{6}n-\frac{31}{12}$$

Subtract \frac{31}{12} from \frac{11}{3} to get \frac{13}{12}.

$$\frac{13}{12}-\frac{1}{6}n$$

Expand

$$-\frac{n}{6}+\frac{13}{12}$$

Solution Steps

Add \frac{2}{3} and 3 to get \frac{11}{3}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n-1\right)^{2}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n^{2}-2n+1\right)+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{1}{4} by n^{2}-2n+1.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}\left(n-1\right)+3\right)$$

Use the distributive property to multiply \frac{2}{3} by n-1.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}n-\frac{2}{3}+3\right)$$

Combine -\frac{1}{2}n and \frac{2}{3}n to get \frac{1}{6}n.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{1}{4}-\frac{2}{3}+3\right)$$

Subtract \frac{2}{3} from \frac{1}{4} to get -\frac{5}{12}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n-\frac{5}{12}+3\right)$$

Add -\frac{5}{12} and 3 to get \frac{31}{12}.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}\right)$$

To find the opposite of \frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}, find the opposite of each term.

$$\frac{1}{4}n^{2}+\frac{11}{3}-\frac{1}{4}n^{2}-\frac{1}{6}n-\frac{31}{12}$$

Combine \frac{1}{4}n^{2} and -\frac{1}{4}n^{2} to get 0.

$$\frac{11}{3}-\frac{1}{6}n-\frac{31}{12}$$

Subtract \frac{31}{12} from \frac{11}{3} to get \frac{13}{12}.

$$\frac{13}{12}-\frac{1}{6}n$$