簡約 (k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3))

解

簡約

\frac{k ( k + 3 )}{4 ( k + 1 ) ( k + 2 )} + \frac{1}{( k + 1 ) ( k + 2 ) ( k + 3 )}

\frac{( k + 1 ) ( k + 4 )}{4 ( k + 2 ) ( k + 3 )}

解答ステップ

\frac{k ( k + 3 )}{4 ( k + 1 ) ( k + 2 )} + \frac{1}{( k + 1 ) ( k + 2 ) ( k + 3 )}

以下の最小公倍数:

4 (k + 1) (k + 2), (k + 1) (k + 2) (k + 3) : 4 (k + 1) (k + 2) (k + 3)

LCMに基づいて分数を調整する

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

分数の規則を適用する:

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

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

簡素化

k (k + 3)^{2} + 4 : k^{3} + 6 k^{2} + 9 k + 4

= \frac{k ^{3} + 6 k ^{2} + 9 k + 4}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )}

キャンセル

\frac{k ^{3} + 6 k ^{2} + 9 k + 4}{4 ( k + 1 ) ( k + 2 ) ( k + 3 )} : \frac{( k + 1 ) ( k + 4 )}{4 ( k + 2 ) ( k + 3 )}

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