展开 (((k+1)(k+2))/2)^2

解答

展开

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

\frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 3 k + 1

求解步骤

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

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

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

(k^{2} + 3 k + 2)^{2} = k^{4} + 6 k^{3} + 13 k^{2} + 12 k + 4

= \frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{2 ^{2}}

2^{2} = 4

= \frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4}

使用分式法则:

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

\frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4} = \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + \frac{4}{4}

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + \frac{4}{4}

使用法则

\frac{a}{a} = 1

\frac{4}{4} = 1

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + 1

消掉

\frac{6 k ^{3}}{4} : \frac{3 k ^{3}}{2}

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + 1

数字相除:

\frac{12}{4} = 3

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 3 k + 1

e x p an d \frac{k ^{2} ( k + 1 ) ^{2}}{4}

s im pl i f y (x + 3 - 2) (x + 3 + 2)

s im pl i f y (x - 1 - 2) (x - 1 + 2)

e x p an d (1 - e^{x})^{2}

s im pl i f y (5 y^{2} - 10 z^{2})^{2}