What is expand (2(k+1)(k+2)(2k+2))/3 ?

The solution to expand (2(k+1)(k+2)(2k+2))/3 is (4k^3)/3+(16k^2)/3+(20k)/3+8/3

Popular >

expand (2(k+1)(k+2)(2k+2))/3

expand

\frac{2 ( k + 1 ) ( k + 2 ) ( 2 k + 2 )}{3}

\frac{4 k ^{3}}{3} + \frac{16 k ^{2}}{3} + \frac{20 k}{3} + \frac{8}{3}

Solution steps

\frac{2 ( k + 1 ) ( k + 2 ) ( 2 k + 2 )}{3}

Expand

2 (k + 1) (k + 2) (2 k + 2) : 4 k^{3} + 16 k^{2} + 20 k + 8

= \frac{4 k ^{3} + 16 k ^{2} + 20 k + 8}{3}

Apply the fraction rule:

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

\frac{4 k ^{3} + 16 k ^{2} + 20 k + 8}{3} = \frac{4 k ^{3}}{3} + \frac{16 k ^{2}}{3} + \frac{20 k}{3} + \frac{8}{3}

= \frac{4 k ^{3}}{3} + \frac{16 k ^{2}}{3} + \frac{20 k}{3} + \frac{8}{3}

\frac{K ( s ^{2} + 4 s + 8 )}{s ^{3} + s ^{2}}

\frac{( x - 6 ) ^{6}}{x}

\frac{( a - 1 ) ^{2}}{4}

2 sin (\frac{π}{3}) \cdot cos (\frac{π}{3}) - 1

\frac{x ^{2}}{( z - 1 ) ( z + 0.5 )}

What is expand (2(k+1)(k+2)(2k+2))/3 ?
The solution to expand (2(k+1)(k+2)(2k+2))/3 is (4k^3)/3+(16k^2)/3+(20k)/3+8/3