What is expand ((n+1)^2*(n+2)^2)/4 ?

The solution to expand ((n+1)^2*(n+2)^2)/4 is (n^4)/4+(3n^3)/2+(13n^2)/4+3n+1

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expand ((n+1)^2*(n+2)^2)/4

expand

\frac{( n + 1 ) ^{2} \cdot ( n + 2 ) ^{2}}{4}

\frac{n ^{4}}{4} + \frac{3 n ^{3}}{2} + \frac{13 n ^{2}}{4} + 3 n + 1

Solution steps

\frac{( n + 1 ) ^{2} ( n + 2 ) ^{2}}{4}

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

= \frac{( n ^{2} + 2 n + 1 ) ( n + 2 ) ^{2}}{4}

(n + 2)^{2} = n^{2} + 4 n + 4

= \frac{( n ^{2} + 2 n + 1 ) ( n ^{2} + 4 n + 4 )}{4}

Expand

(n^{2} + 2 n + 1) (n^{2} + 4 n + 4) : n^{4} + 6 n^{3} + 13 n^{2} + 12 n + 4

= \frac{n ^{4} + 6 n ^{3} + 13 n ^{2} + 12 n + 4}{4}

Apply the fraction rule:

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

\frac{n ^{4} + 6 n ^{3} + 13 n ^{2} + 12 n + 4}{4} = \frac{n ^{4}}{4} + \frac{6 n ^{3}}{4} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + \frac{4}{4}

= \frac{n ^{4}}{4} + \frac{6 n ^{3}}{4} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + \frac{4}{4}

Apply rule

\frac{a}{a} = 1

\frac{4}{4} = 1

= \frac{n ^{4}}{4} + \frac{6 n ^{3}}{4} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + 1

Cancel

\frac{6 n ^{3}}{4} : \frac{3 n ^{3}}{2}

= \frac{n ^{4}}{4} + \frac{3 n ^{3}}{2} + \frac{13 n ^{2}}{4} + \frac{12 n}{4} + 1

Divide the numbers:

\frac{12}{4} = 3

= \frac{n ^{4}}{4} + \frac{3 n ^{3}}{2} + \frac{13 n ^{2}}{4} + 3 n + 1

\frac{2 s}{( s ^{2} + 4 s + 5 ) ^{2}}

\frac{4 s + 3}{6 s ( s - 1 )}

(3 x + π^{3}) (2 x - 7)

(x^{3} - x + 1) l e f t (x^{2} - x)

\frac{2 ( 1 - i ) ^{5}}{8}

What is expand ((n+1)^2*(n+2)^2)/4 ?
The solution to expand ((n+1)^2*(n+2)^2)/4 is (n^4)/4+(3n^3)/2+(13n^2)/4+3n+1