What is expand (n(n+2)(n+3)(n+4))/4 ?

The solution to expand (n(n+2)(n+3)(n+4))/4 is (n^4)/4+(9n^3)/4+(13n^2)/2+6n

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expand (n(n+2)(n+3)(n+4))/4

expand

\frac{n ( n + 2 ) ( n + 3 ) ( n + 4 )}{4}

\frac{n ^{4}}{4} + \frac{9 n ^{3}}{4} + \frac{13 n ^{2}}{2} + 6 n

Solution steps

\frac{n ( n + 2 ) ( n + 3 ) ( n + 4 )}{4}

Expand

n (n + 2) (n + 3) (n + 4) : n^{4} + 9 n^{3} + 26 n^{2} + 24 n

= \frac{n ^{4} + 9 n ^{3} + 26 n ^{2} + 24 n}{4}

Apply the fraction rule:

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

\frac{n ^{4} + 9 n ^{3} + 26 n ^{2} + 24 n}{4} = \frac{n ^{4}}{4} + \frac{9 n ^{3}}{4} + \frac{26 n ^{2}}{4} + \frac{24 n}{4}

= \frac{n ^{4}}{4} + \frac{9 n ^{3}}{4} + \frac{26 n ^{2}}{4} + \frac{24 n}{4}

Cancel

\frac{26 n ^{2}}{4} : \frac{13 n ^{2}}{2}

= \frac{n ^{4}}{4} + \frac{9 n ^{3}}{4} + \frac{13 n ^{2}}{2} + \frac{24 n}{4}

Divide the numbers:

\frac{24}{4} = 6

= \frac{n ^{4}}{4} + \frac{9 n ^{3}}{4} + \frac{13 n ^{2}}{2} + 6 n

y (- e^{- y} = x^{3} + c)

\frac{5 s + 3}{( s - 1 ) ( s ^{2} + 2 s + 5 )}

e^{(50 - t)^{2}} - 2 (50 - t) + c

(5)^{0}

(x + \frac{1}{x}) d x

What is expand (n(n+2)(n+3)(n+4))/4 ?
The solution to expand (n(n+2)(n+3)(n+4))/4 is (n^4)/4+(9n^3)/4+(13n^2)/2+6n