Solve =2+1/2*1/6(n-1)n(2n+1)+1/2(3/2+frac{1}{2}n+frac{1}{2})(n-1)

$$= 2 + \frac { 1 } { 2 } \times \frac { 1 } { 6 } ( n - 1 ) n ( 2 n + 1 ) + \frac { 1 } { 2 } ( \frac { 3 } { 2 } + \frac { 1 } { 2 } n + \frac { 1 } { 2 } ) ( n - 1 )$$

Evaluate

$$\frac{n^{3}+n^{2}+4n+6}{6}$$

Solution Steps

Multiply \frac{1}{2} times \frac{1}{6} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\times 1}{2\times 6}.

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

Use the distributive property to multiply \frac{1}{12} by n-1.

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

Multiply \frac{1}{12} and -1 to get -\frac{1}{12}.

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

Use the distributive property to multiply \frac{1}{12}n-\frac{1}{12} by n.

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

Multiply n and n to get n^{2}.

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

Apply the distributive property by multiplying each term of \frac{1}{12}n^{2}-\frac{1}{12}n by each term of 2n+1.

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

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Multiply n and n to get n^{2}.

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

Multiply \frac{1}{12} and 2 to get \frac{2}{12}.

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

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

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

Express -\frac{1}{12}\times 2 as a single fraction.

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

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

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

Combine \frac{1}{12}n^{2} and -\frac{1}{6}n^{2} to get -\frac{1}{12}n^{2}.

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

Since \frac{3}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Add 3 and 1 to get 4.

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

Divide 4 by 2 to get 2.

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

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

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

Cancel out 2 and 2.

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

Multiply \frac{1}{2} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\times 1}{2\times 2}.

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

Apply the distributive property by multiplying each term of 1+\frac{1}{4}n by each term of n-1.

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

Multiply n and n to get n^{2}.

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

Multiply \frac{1}{4} and -1 to get -\frac{1}{4}.

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

Combine n and -\frac{1}{4}n to get \frac{3}{4}n.

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

Combine -\frac{1}{12}n and \frac{3}{4}n to get \frac{2}{3}n.

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

Subtract 1 from 2 to get 1.

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

Combine -\frac{1}{12}n^{2} and \frac{1}{4}n^{2} to get \frac{1}{6}n^{2}.

$$1+\frac{1}{6}n^{3}+\frac{1}{6}n^{2}+\frac{2}{3}n$$

Expand

$$\frac{n^{3}}{6}+\frac{n^{2}}{6}+\frac{2n}{3}+1$$

Solution Steps

Multiply \frac{1}{2} times \frac{1}{6} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\times 1}{2\times 6}.

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

Use the distributive property to multiply \frac{1}{12} by n-1.

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

Multiply \frac{1}{12} and -1 to get -\frac{1}{12}.

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

Use the distributive property to multiply \frac{1}{12}n-\frac{1}{12} by n.

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

Multiply n and n to get n^{2}.

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

Apply the distributive property by multiplying each term of \frac{1}{12}n^{2}-\frac{1}{12}n by each term of 2n+1.

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

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Multiply n and n to get n^{2}.

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

Multiply \frac{1}{12} and 2 to get \frac{2}{12}.

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

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

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

Express -\frac{1}{12}\times 2 as a single fraction.

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

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

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

Combine \frac{1}{12}n^{2} and -\frac{1}{6}n^{2} to get -\frac{1}{12}n^{2}.

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

Since \frac{3}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Add 3 and 1 to get 4.

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

Divide 4 by 2 to get 2.

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

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

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

Cancel out 2 and 2.

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

Multiply \frac{1}{2} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\times 1}{2\times 2}.

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

Apply the distributive property by multiplying each term of 1+\frac{1}{4}n by each term of n-1.

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

Multiply n and n to get n^{2}.

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

Multiply \frac{1}{4} and -1 to get -\frac{1}{4}.

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

Combine n and -\frac{1}{4}n to get \frac{3}{4}n.

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

Combine -\frac{1}{12}n and \frac{3}{4}n to get \frac{2}{3}n.

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

Subtract 1 from 2 to get 1.

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

Combine -\frac{1}{12}n^{2} and \frac{1}{4}n^{2} to get \frac{1}{6}n^{2}.

$$1+\frac{1}{6}n^{3}+\frac{1}{6}n^{2}+\frac{2}{3}n$$