What is expand ((n+1)(2n+1)(2n+3))/3 ?

The solution to expand ((n+1)(2n+1)(2n+3))/3 is (4n^3)/3+4n^2+(11n)/3+1

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expand ((n+1)(2n+1)(2n+3))/3

expand

\frac{( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 )}{3}

\frac{4 n ^{3}}{3} + 4 n^{2} + \frac{11 n}{3} + 1

Solution steps

\frac{( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 )}{3}

Expand

(n + 1) (2 n + 1) (2 n + 3) : 4 n^{3} + 12 n^{2} + 11 n + 3

= \frac{4 n ^{3} + 12 n ^{2} + 11 n + 3}{3}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{4 n ^{3} + 12 n ^{2} + 11 n + 3}{3} = \frac{4 n ^{3}}{3} + \frac{12 n ^{2}}{3} + \frac{11 n}{3} + \frac{3}{3}

= \frac{4 n ^{3}}{3} + \frac{12 n ^{2}}{3} + \frac{11 n}{3} + \frac{3}{3}

Apply rule

\frac{a}{a} = 1

\frac{3}{3} = 1

= \frac{4 n ^{3}}{3} + \frac{12 n ^{2}}{3} + \frac{11 n}{3} + 1

Divide the numbers:

\frac{12}{3} = 4

= \frac{4 n ^{3}}{3} + 4 n^{2} + \frac{11 n}{3} + 1

\frac{x ^{n + 1} - ( n + 1 ) x + n}{x - 1}

\frac{( x + h ) ^{2} - 1}{( x + h ) + 4}

(e^{t} - e^{- t})^{3}

(2 x + 2Δ x + 3) (4 x - 5)

\frac{( x + 4 ) ( x + 8 )}{x + 5}

What is expand ((n+1)(2n+1)(2n+3))/3 ?
The solution to expand ((n+1)(2n+1)(2n+3))/3 is (4n^3)/3+4n^2+(11n)/3+1