\[ \ \ \ \ \dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}\\ =\dfrac{1}{1 \times 2}+\dfrac{1}{2 \times 3}+\dfrac{1}{3 \times 4}+\dfrac{1}{4 \times 5}\\ =\left(\dfrac{1}{1}-\dfrac{1}{2}\right) +\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\\ \ \ \ \ +\left(\dfrac{1}{3}-\dfrac{1}{4}\right) +\left(\dfrac{1}{4}-\dfrac{1}{5}\right)\\ =\dfrac{1}{1}-\dfrac{1}{5}\\ =\dfrac{4}{5} \]
分数を「二つの分数の差」にすることを部分分数分解といいます。
上の例では $\dfrac{1}{6}$ を $\dfrac{1}{2}-\dfrac{1}{3}$ にしています。
部分分数分解の公式
\[ \dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1} \]
例をあげてみます。
[
\dfrac{1}{2}=\dfrac{1}{1}-\dfrac{1}{2}
]
[
\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}
]
[
\dfrac{1}{12}=\dfrac{1}{3}-\dfrac{1}{4}
]
[
\dfrac{1}{20}=\dfrac{1}{4}-\dfrac{1}{5}
]
[
\dfrac{1}{30}=\dfrac{1}{5}-\dfrac{1}{6}
]
[
\dfrac{1}{42}=\dfrac{1}{6}-\dfrac{1}{7}
]
[
\dfrac{1}{56}=\dfrac{1}{7}-\dfrac{1}{8}
]
部分分数分解の問題
部分分数分解を理解できたか次の問題でチェックしてみよう。
[
\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}\
+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}
]
解答
[
\ \ \ \ \dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\
=\dfrac{1}{1 \times 2}+\dfrac{1}{2 \times 3}+\dfrac{1}{3 \times 4}\
\ \ \ \ +\dfrac{1}{4 \times 5}+\dfrac{1}{5 \times 6}+\dfrac{1}{6 \times 7}\
=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)
+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\
\ \ \ \ +\left(\dfrac{1}{3}-\dfrac{1}{4}\right)
+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)\
\ \ \ \ +\left(\dfrac{1}{5}-\dfrac{1}{6}\right)
+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)\
=\dfrac{1}{1}-\dfrac{1}{7}\
=\dfrac{6}{7}
]