What is expand (k(k+1)(2k+1))/6+(k+1) ?

The solution to expand (k(k+1)(2k+1))/6+(k+1) is (k^3)/3+(k^2)/2+k/6+k+1

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expand (k(k+1)(2k+1))/6+(k+1)

expand

\frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)

\frac{k ^{3}}{3} + \frac{k ^{2}}{2} + \frac{k}{6} + k + 1

Solution steps

\frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)

Remove parentheses:

(a) = a

= \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + k + 1

Expand

k (k + 1) (2 k + 1) : 2 k^{3} + 3 k^{2} + k

= \frac{2 k ^{3} + 3 k ^{2} + k}{6} + k + 1

Apply the fraction rule:

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

\frac{2 k ^{3} + 3 k ^{2} + k}{6} = \frac{2 k ^{3}}{6} + \frac{3 k ^{2}}{6} + \frac{k}{6}

= \frac{2 k ^{3}}{6} + \frac{3 k ^{2}}{6} + \frac{k}{6} + k + 1

Cancel

\frac{2 k ^{3}}{6} : \frac{k ^{3}}{3}

= \frac{k ^{3}}{3} + \frac{3 k ^{2}}{6} + \frac{k}{6} + k + 1

Cancel

\frac{3 k ^{2}}{6} : \frac{k ^{2}}{2}

= \frac{k ^{3}}{3} + \frac{k ^{2}}{2} + \frac{k}{6} + k + 1

(e^{- x} - x e^{- x})^{2}

\frac{5 x}{( x + 1 ) ^{2}}

\frac{( x + y ) ^{2}}{2}

\frac{5 ( f - 32 )}{9}

(- 3, 2) y (7, - 3)

What is expand (k(k+1)(2k+1))/6+(k+1) ?
The solution to expand (k(k+1)(2k+1))/6+(k+1) is (k^3)/3+(k^2)/2+k/6+k+1