展开 ((k+1)^2(k+2)^2(2k^2+6k+3))/(12)

解答

展开

\frac{( k + 1 ) ^{2} ( k + 2 ) ^{2} ( 2 k ^{2} + 6 k + 3 )}{12}

\frac{k ^{6}}{6} + \frac{3 k ^{5}}{2} + \frac{65 k ^{4}}{12} + 10 k^{3} + \frac{119 k ^{2}}{12} + 5 k + 1

求解步骤

\frac{( k + 1 ) ^{2} ( k + 2 ) ^{2} ( 2 k ^{2} + 6 k + 3 )}{12}

(k + 1)^{2} = k^{2} + 2 k + 1

= \frac{( k ^{2} + 2 k + 1 ) ( k + 2 ) ^{2} ( 2 k ^{2} + 6 k + 3 )}{12}

(k + 2)^{2} = k^{2} + 4 k + 4

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

乘开

(k^{2} + 2 k + 1) (k^{2} + 4 k + 4) (2 k^{2} + 6 k + 3) : 2 k^{6} + 18 k^{5} + 65 k^{4} + 120 k^{3} + 119 k^{2} + 60 k + 12

= \frac{2 k ^{6} + 18 k ^{5} + 65 k ^{4} + 120 k ^{3} + 119 k ^{2} + 60 k + 12}{12}

使用分式法则:

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

\frac{2 k ^{6} + 18 k ^{5} + 65 k ^{4} + 120 k ^{3} + 119 k ^{2} + 60 k + 12}{12} = \frac{2 k ^{6}}{12} + \frac{18 k ^{5}}{12} + \frac{65 k ^{4}}{12} + \frac{120 k ^{3}}{12} + \frac{119 k ^{2}}{12} + \frac{60 k}{12} + \frac{12}{12}

= \frac{2 k ^{6}}{12} + \frac{18 k ^{5}}{12} + \frac{65 k ^{4}}{12} + \frac{120 k ^{3}}{12} + \frac{119 k ^{2}}{12} + \frac{60 k}{12} + \frac{12}{12}

使用法则

\frac{a}{a} = 1

\frac{12}{12} = 1

= \frac{2 k ^{6}}{12} + \frac{18 k ^{5}}{12} + \frac{65 k ^{4}}{12} + \frac{120 k ^{3}}{12} + \frac{119 k ^{2}}{12} + \frac{60 k}{12} + 1

\frac{2 k ^{6}}{12} = \frac{k ^{6}}{6}

\frac{18 k ^{5}}{12} = \frac{3 k ^{5}}{2}

\frac{120 k ^{3}}{12} = 10 k^{3}

\frac{60 k}{12} = 5 k

= \frac{k ^{6}}{6} + \frac{3 k ^{5}}{2} + \frac{65 k ^{4}}{12} + 10 k^{3} + \frac{119 k ^{2}}{12} + 5 k + 1

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