Solve 1/4n^2+2/3+3-1/4left(n-1right)^2+2/3left(n-1right)+3

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

Evaluate

$$\frac{7n}{6}+\frac{23}{4}$$

Solution Steps

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

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

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

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

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

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

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

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

Subtract \frac{1}{4} from \frac{11}{3} to get \frac{41}{12}.

$$\frac{41}{12}+\frac{1}{2}n+\frac{2}{3}\left(n-1\right)+3$$

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

$$\frac{41}{12}+\frac{1}{2}n+\frac{2}{3}n-\frac{2}{3}+3$$

Combine \frac{1}{2}n and \frac{2}{3}n to get \frac{7}{6}n.

$$\frac{41}{12}+\frac{7}{6}n-\frac{2}{3}+3$$

Subtract \frac{2}{3} from \frac{41}{12} to get \frac{11}{4}.

$$\frac{11}{4}+\frac{7}{6}n+3$$

Add \frac{11}{4} and 3 to get \frac{23}{4}.

$$\frac{23}{4}+\frac{7}{6}n$$

Expand

$$\frac{7n}{6}+\frac{23}{4}$$

Solution Steps

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

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

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

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

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

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

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

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

Subtract \frac{1}{4} from \frac{11}{3} to get \frac{41}{12}.

$$\frac{41}{12}+\frac{1}{2}n+\frac{2}{3}\left(n-1\right)+3$$

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

$$\frac{41}{12}+\frac{1}{2}n+\frac{2}{3}n-\frac{2}{3}+3$$

Combine \frac{1}{2}n and \frac{2}{3}n to get \frac{7}{6}n.

$$\frac{41}{12}+\frac{7}{6}n-\frac{2}{3}+3$$

Subtract \frac{2}{3} from \frac{41}{12} to get \frac{11}{4}.

$$\frac{11}{4}+\frac{7}{6}n+3$$

Add \frac{11}{4} and 3 to get \frac{23}{4}.

$$\frac{23}{4}+\frac{7}{6}n$$