What is expand (n^2(n+1)^2(2n^2+2n-1))/(12) ?

The solution to expand (n^2(n+1)^2(2n^2+2n-1))/(12) is (n^6)/6+(n^5)/2+(5n^4)/(12)-(n^2)/(12)

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expand (n^2(n+1)^2(2n^2+2n-1))/(12)

expand

\frac{n ^{2} ( n + 1 ) ^{2} ( 2 n ^{2} + 2 n - 1 )}{12}

\frac{n ^{6}}{6} + \frac{n ^{5}}{2} + \frac{5 n ^{4}}{12} - \frac{n ^{2}}{12}

Solution steps

\frac{n ^{2} ( n + 1 ) ^{2} ( 2 n ^{2} + 2 n - 1 )}{12}

(n + 1)^{2} = n^{2} + 2 n + 1

= \frac{n ^{2} ( n ^{2} + 2 n + 1 ) ( 2 n ^{2} + 2 n - 1 )}{12}

Expand

n^{2} (n^{2} + 2 n + 1) (2 n^{2} + 2 n - 1) : 2 n^{6} + 6 n^{5} + 5 n^{4} - n^{2}

= \frac{2 n ^{6} + 6 n ^{5} + 5 n ^{4} - n ^{2}}{12}

Apply the fraction rule:

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

\frac{2 n ^{6} + 6 n ^{5} + 5 n ^{4} - n ^{2}}{12} = \frac{2 n ^{6}}{12} + \frac{6 n ^{5}}{12} + \frac{5 n ^{4}}{12} - \frac{n ^{2}}{12}

= \frac{2 n ^{6}}{12} + \frac{6 n ^{5}}{12} + \frac{5 n ^{4}}{12} - \frac{n ^{2}}{12}

Cancel

\frac{2 n ^{6}}{12} : \frac{n ^{6}}{6}

= \frac{n ^{6}}{6} + \frac{6 n ^{5}}{12} + \frac{5 n ^{4}}{12} - \frac{n ^{2}}{12}

Cancel

\frac{6 n ^{5}}{12} : \frac{n ^{5}}{2}

= \frac{n ^{6}}{6} + \frac{n ^{5}}{2} + \frac{5 n ^{4}}{12} - \frac{n ^{2}}{12}

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What is expand (n^2(n+1)^2(2n^2+2n-1))/(12) ?
The solution to expand (n^2(n+1)^2(2n^2+2n-1))/(12) is (n^6)/6+(n^5)/2+(5n^4)/(12)-(n^2)/(12)