What is expand ((n+1)^4+2(n+1)^3+(n+1)^2)/4 ?

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

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expand ((n+1)^4+2(n+1)^3+(n+1)^2)/4

expand

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

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

Solution steps

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

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

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

Expand

(n + 1)^{4} + 2 (n^{3} + 3 n^{2} + 3 n + 1) + (n + 1)^{2} : 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{\partial}{\partial x} ((z) (e^{x yz} - x - y))

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

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

- \frac{1}{8} (x + 2)^{2} (x - 4)^{2}

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

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