(1)
2/{1・3}=[1/1]-[1/3]
2/{3・5}=[1/3]-[1/5]
2/{5・7}=[1/5]-[1/7]
2/{7・9}=[1/7]-[1/9]
・・・・・・
2/{(2n-1)(2n+1)}=[1/{(2n-1)]-[1/(2n+1)]
という感じの変形の後、和を考えます
●和=[1/1]-[1/(2n+1)]=[2n/(2n+1)]
分子が2の場合の解き方が分かりません
1なら分かるのですが...
(1)
2/{1・3}=[1/1]-[1/3]
2/{3・5}=[1/3]-[1/5]
2/{5・7}=[1/5]-[1/7]
2/{7・9}=[1/7]-[1/9]
・・・・・・
2/{(2n-1)(2n+1)}=[1/{(2n-1)]-[1/(2n+1)]
という感じの変形の後、和を考えます
●和=[1/1]-[1/(2n+1)]=[2n/(2n+1)]
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