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

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

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

expand

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

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

Solution steps

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

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

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

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

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

Expand

n^{2} (n^{2} + 2 n + 1) + 4 (n^{3} + 3 n^{2} + 3 n + 1) : 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{x ( 4 - x )}{2}

x (2 x^{3} - 3 y^{3}) d y - y (2 x^{3} - y^{3}) d x

P (- 2, 4) y Q (4, - 6)

2 π (\frac{4 π}{3} - 1)

(n + 1)! (n + 2)

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