What is expand (n(n+1)(2+n^2+n))/8+(n(n+1))/2 ?

The solution to expand (n(n+1)(2+n^2+n))/8+(n(n+1))/2 is (n^4)/8+(n^3)/4+(7n^2)/8+(3n)/4

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expand (n(n+1)(2+n^2+n))/8+(n(n+1))/2

expand

\frac{n ( n + 1 ) ( 2 + n ^{2} + n )}{8} + \frac{n ( n + 1 )}{2}

\frac{n ^{4}}{8} + \frac{n ^{3}}{4} + \frac{7 n ^{2}}{8} + \frac{3 n}{4}

Solution steps

\frac{n ( n + 1 ) ( 2 + n ^{2} + n )}{8} + \frac{n ( n + 1 )}{2}

Expand

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

= \frac{n ^{4} + 2 n ^{3} + 3 n ^{2} + 2 n}{8} + \frac{n ( n + 1 )}{2}

Expand

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

= \frac{n ^{4} + 2 n ^{3} + 3 n ^{2} + 2 n}{8} + \frac{n ^{2} + n}{2}

Least Common Multiplier of

8, 2 : 8

Adjust Fractions based on the LCM

= \frac{n ^{4} + 2 n ^{3} + 3 n ^{2} + 2 n}{8} + \frac{( n ^{2} + n ) \cdot 4}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{n ^{4} + 2 n ^{3} + 3 n ^{2} + 2 n + ( n ^{2} + n ) \cdot 4}{8}

Expand

n^{4} + 2 n^{3} + 3 n^{2} + 2 n + (n^{2} + n) \cdot 4 : n^{4} + 2 n^{3} + 7 n^{2} + 6 n

= \frac{n ^{4} + 2 n ^{3} + 7 n ^{2} + 6 n}{8}

Apply the fraction rule:

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

\frac{n ^{4} + 2 n ^{3} + 7 n ^{2} + 6 n}{8} = \frac{n ^{4}}{8} + \frac{2 n ^{3}}{8} + \frac{7 n ^{2}}{8} + \frac{6 n}{8}

= \frac{n ^{4}}{8} + \frac{2 n ^{3}}{8} + \frac{7 n ^{2}}{8} + \frac{6 n}{8}

Cancel

\frac{2 n ^{3}}{8} : \frac{n ^{3}}{4}

= \frac{n ^{4}}{8} + \frac{n ^{3}}{4} + \frac{7 n ^{2}}{8} + \frac{6 n}{8}

Cancel

\frac{6 n}{8} : \frac{3 n}{4}

= \frac{n ^{4}}{8} + \frac{n ^{3}}{4} + \frac{7 n ^{2}}{8} + \frac{3 n}{4}

\frac{( sec ( x ) tan ( x ) )}{( 1 + sec ^{(2)} ( x ) ) d x}

4 (5 - x = 4)

\frac{\partial}{\partial x} ((u (x, y)) (2 x + 3 x y))

\frac{\partial}{\partial x} ((y) (y + sin (y) - x))

\frac{2 e ^{- 3 s}}{s ^{2} ( s - 1 ) ( s - 2 )}

What is expand (n(n+1)(2+n^2+n))/8+(n(n+1))/2 ?
The solution to expand (n(n+1)(2+n^2+n))/8+(n(n+1))/2 is (n^4)/8+(n^3)/4+(7n^2)/8+(3n)/4