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

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

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

expand

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

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

Solution steps

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

\frac{( n + 1 ) ( n + 2 )}{2} = \frac{n ^{2} + 3 n + 2}{2}

= (\frac{n ^{2} + 3 n + 2}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( n ^{2} + 3 n + 2 ) ^{2}}{2 ^{2}}

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

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

2^{2} = 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

x^{\frac{1}{2}} (x^{\frac{3}{2}} - 4) (x^{\frac{1}{2}} + 4)

(1 - ln (x)) d y

x + 4 (x - 4 + 3 x - 4)

(- \frac{1}{23 , 2 + 0 , 9 K}) \cdot (- 0.9 K^{2} + 53.6 K - 793.6)

\frac{1}{( z + 1 ) ( z + 2 )}

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