设an=1+1/2+1/3+...1/(3n-1),(n∈N+)则a(n+1)-an等于
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设an=1+1/2+1/3+...1/(3n-1),(n∈N+)则a(n+1)-an等于设an=1+1/2+1/3+...1/(3n-1),(n∈N+)则a(n+1)-an等于设an=1+1/2+1/
设an=1+1/2+1/3+...1/(3n-1),(n∈N+)则a(n+1)-an等于
设an=1+1/2+1/3+...1/(3n-1),(n∈N+)则a(n+1)-an等于
设an=1+1/2+1/3+...1/(3n-1),(n∈N+)则a(n+1)-an等于
注意到an=1+1/2+1/3+...+1/(3n-1)
它的分母是连续自然数
所以a(n+1)=1+1/2+1/3+...+1/(3n-1)+1/(3n)+1/(3n+1)+1/(3n+2)
∴a(n+1)-an=1/(3n)+1/(3n+1)+1/(3n+2)
这种很简单的,根据an=1+1/2+1/3+...1/(3n-1)
可以知道a(n+1)=1+1/2+1/3+...1/(3n-1)+1/[3(n+1]-1] (将n换成n+1就知道最后一项是什么,这种题都这么做)
所以a(n+1)-an=1/[3(n+1]-1]=1/(3n-2)
1/(3n+2)
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