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受欢迎的代数 >

展开 (((k+1)+1)(2(k+1)+1)(2(k+1)+3))/3

解答

展开

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

解答

\frac{4 k ^{3}}{3} + 8 k^{2} + \frac{47 k}{3} + 10

求解步骤

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

去除括号:

(a) = a

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

数字相加:

1 + 1 = 2

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

乘开

2 (k + 1) + 1 : 2 k + 3

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

乘开

2 (k + 1) + 3 : 2 k + 5

= \frac{( k + 2 ) ( 2 k + 3 ) ( 2 k + 5 )}{3}

乘开

(k + 2) (2 k + 3) (2 k + 5) : 4 k^{3} + 24 k^{2} + 47 k + 30

= \frac{4 k ^{3} + 24 k ^{2} + 47 k + 30}{3}

使用分式法则:

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

\frac{4 k ^{3} + 24 k ^{2} + 47 k + 30}{3} = \frac{4 k ^{3}}{3} + \frac{24 k ^{2}}{3} + \frac{47 k}{3} + \frac{30}{3}

= \frac{4 k ^{3}}{3} + \frac{24 k ^{2}}{3} + \frac{47 k}{3} + \frac{30}{3}

数字相除:

\frac{24}{3} = 8

= \frac{4 k ^{3}}{3} + 8 k^{2} + \frac{47 k}{3} + \frac{30}{3}

数字相除:

\frac{30}{3} = 10

= \frac{4 k ^{3}}{3} + 8 k^{2} + \frac{47 k}{3} + 10

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