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

The solution to (k(k+1)(2k+1))/6+(k+1)^2 is (2k^3+9k^2+13k+6)/6

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simplify (k(k+1)(2k+1))/6+(k+1)^2

simplify

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

\frac{2 k ^{3} + 9 k ^{2} + 13 k + 6}{6}

Solution steps

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

Apply the fraction rule:

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

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

Simplify

(k + 1)^{2} \cdot 6 + k (k + 1) (2 k + 1) : 2 k^{3} + 9 k^{2} + 13 k + 6

= \frac{2 k ^{3} + 9 k ^{2} + 13 k + 6}{6}

- \frac{2}{5} - \frac{4}{5} + (- \frac{9}{10})

- 1 < x < 6

- (\frac{5}{3})^{4}

x^{2} + 2 x > 3

\frac{- 3 ( 5 - x )}{10} - \frac{3 x}{2} = 7 - \frac{5 x}{3}

What is (k(k+1)(2k+1))/6+(k+1)^2 ?
The solution to (k(k+1)(2k+1))/6+(k+1)^2 is (2k^3+9k^2+13k+6)/6