What is expand (n(n+1)(2n+1)(3n+1))/(24) ?

The solution to expand (n(n+1)(2n+1)(3n+1))/(24) is (n^4)/4+(11n^3}{24}+\frac{n^2)/4+n/(24)

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expand (n(n+1)(2n+1)(3n+1))/(24)

expand

\frac{n ( n + 1 ) ( 2 n + 1 ) ( 3 n + 1 )}{24}

\frac{n ^{4}}{4} + \frac{11 n ^{3}}{24} + \frac{n ^{2}}{4} + \frac{n}{24}

Solution steps

\frac{n ( n + 1 ) ( 2 n + 1 ) ( 3 n + 1 )}{24}

Expand

n (n + 1) (2 n + 1) (3 n + 1) : 6 n^{4} + 11 n^{3} + 6 n^{2} + n

= \frac{6 n ^{4} + 11 n ^{3} + 6 n ^{2} + n}{24}

Apply the fraction rule:

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

\frac{6 n ^{4} + 11 n ^{3} + 6 n ^{2} + n}{24} = \frac{6 n ^{4}}{24} + \frac{11 n ^{3}}{24} + \frac{6 n ^{2}}{24} + \frac{n}{24}

= \frac{6 n ^{4}}{24} + \frac{11 n ^{3}}{24} + \frac{6 n ^{2}}{24} + \frac{n}{24}

Cancel

\frac{6 n ^{4}}{24} : \frac{n ^{4}}{4}

= \frac{n ^{4}}{4} + \frac{11 n ^{3}}{24} + \frac{6 n ^{2}}{24} + \frac{n}{24}

Cancel

\frac{6 n ^{2}}{24} : \frac{n ^{2}}{4}

= \frac{n ^{4}}{4} + \frac{11 n ^{3}}{24} + \frac{n ^{2}}{4} + \frac{n}{24}

\frac{16}{5 x ( 3 x + 4 )}

\frac{12 ( 7 + 4 7 )}{7}

v^{2} (y d v + v d y)

(1 + 2 cos (θ))^{2}

\frac{π ( ( 16 A ^{2} + 9 ) ^{\frac{3}{2}} - 27 )}{72}

What is expand (n(n+1)(2n+1)(3n+1))/(24) ?
The solution to expand (n(n+1)(2n+1)(3n+1))/(24) is (n^4)/4+(11n^3}{24}+\frac{n^2)/4+n/(24)