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} ( 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

- 2 (5)^{2} + 400 (6) - 2000

f (t) = e^{(50 (\frac{e ^{t} - 1}{t} - 1))}

e^{(- t - X)} \cdot u (t - X)

f (x) = \frac{1}{2} (e^{x} + e^{2 - x})

\frac{( 16 x ^{2} - 1 ) ^{2}}{64 x ^{2}}

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