设f(n)=1/(n+1)+1/(n+2)+...+1/2n,则f(n+1)-f(n)等于()我想f(n+1)与f(n)中间的都消掉了,就剩两边一个1/(2n+1)与1/(n+1)了.所以最后应该是1/(2n+1)-1/(n+1).可是为什么错了.
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![设f(n)=1/(n+1)+1/(n+2)+...+1/2n,则f(n+1)-f(n)等于()我想f(n+1)与f(n)中间的都消掉了,就剩两边一个1/(2n+1)与1/(n+1)了.所以最后应该是1/(2n+1)-1/(n+1).可是为什么错了.](/uploads/image/z/6765790-22-0.jpg?t=%E8%AE%BEf%28n%29%3D1%2F%28n%2B1%29%2B1%2F%28n%2B2%29%2B...%2B1%2F2n%2C%E5%88%99f%EF%BC%88n%2B1%EF%BC%89-f%EF%BC%88n%EF%BC%89%E7%AD%89%E4%BA%8E%EF%BC%88%EF%BC%89%E6%88%91%E6%83%B3f%EF%BC%88n%2B1%EF%BC%89%E4%B8%8Ef%EF%BC%88n%EF%BC%89%E4%B8%AD%E9%97%B4%E7%9A%84%E9%83%BD%E6%B6%88%E6%8E%89%E4%BA%86%2C%E5%B0%B1%E5%89%A9%E4%B8%A4%E8%BE%B9%E4%B8%80%E4%B8%AA1%2F%EF%BC%882n%2B1%EF%BC%89%E4%B8%8E1%2F%28n%2B1%29%E4%BA%86.%E6%89%80%E4%BB%A5%E6%9C%80%E5%90%8E%E5%BA%94%E8%AF%A5%E6%98%AF1%2F%EF%BC%882n%2B1%EF%BC%89-1%2F%28n%2B1%29.%E5%8F%AF%E6%98%AF%E4%B8%BA%E4%BB%80%E4%B9%88%E9%94%99%E4%BA%86.)
设f(n)=1/(n+1)+1/(n+2)+...+1/2n,则f(n+1)-f(n)等于()我想f(n+1)与f(n)中间的都消掉了,就剩两边一个1/(2n+1)与1/(n+1)了.所以最后应该是1/(2n+1)-1/(n+1).可是为什么错了.
设f(n)=1/(n+1)+1/(n+2)+...+1/2n,则f(n+1)-f(n)等于()
我想f(n+1)与f(n)中间的都消掉了,就剩两边一个1/(2n+1)与1/(n+1)了.所以最后应该是1/(2n+1)-1/(n+1).可是为什么错了.
设f(n)=1/(n+1)+1/(n+2)+...+1/2n,则f(n+1)-f(n)等于()我想f(n+1)与f(n)中间的都消掉了,就剩两边一个1/(2n+1)与1/(n+1)了.所以最后应该是1/(2n+1)-1/(n+1).可是为什么错了.
f(n+1)=1/(n+1+1)+1/(n+1+2)+...+1/2(n+1)
所以
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
f(n)=1/(n+1)+1/(n+2)+...+1/(2n)
f(n+1)-f(n)
=1/(n+2)+1/(n+3)+...+1/(2(n+1)) - [1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(2n+1) + 1/(2n+2) -1/(n+1)
=1/(2n+1) - 1/(2n+2)
f(n+1)=1/(n+1+1)+1/(n+1+2)+...+1/2(n+1)
=1/(n+2)+1/(n+3)+...+1/(2n+2)
f(n)=1/(n+1)+1/(n+2)+...+1/2n
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/(2n+1)1/(2n+2)