已知 ∑r=1n1r(r+1)=nn+1\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}。

  1. (a)

    证明 ∑r=2n1r(r−1)=1−1n\displaystyle\sum_{r=2}^{n} \frac{1}{r(r-1)} = 1 - \frac{1}{n}。

  2. (b)

    求 ∑r=2n(12r+1r(r−1))\displaystyle\sum_{r=2}^{n} \left( \frac{1}{2^r} + \frac{1}{r(r-1)} \right)。

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完整解法

3(a)

1. Express the required sum in a form that matches the given identity by changing the index.

Let k=r−1k = r - 1. Then r=k+1r = k + 1.
When r=2r = 2, k=1k = 1.
When r=nr = n, k=n−1k = n - 1.
∑r=2n1r(r−1)=∑k=1n−11(k+1)k=∑k=1n−11k(k+1)\sum_{r=2}^{n} \frac{1}{r(r-1)} = \sum_{k=1}^{n-1} \frac{1}{(k+1)k} = \sum_{k=1}^{n-1} \frac{1}{k(k+1)}

2. Apply the given identity and simplify to obtain the result.

Using the given identity ∑r=1n1r(r+1)=nn+1\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1} by replacing nn with n−1n-1:
∑k=1n−11k(k+1)=n−1(n−1)+1\sum_{k=1}^{n-1} \frac{1}{k(k+1)} = \frac{n-1}{(n-1)+1}
=n−1n= \frac{n-1}{n}
=nn−1n= \frac{n}{n} - \frac{1}{n}
=1−1n (Shown)= 1 - \frac{1}{n} \text{ (Shown)}

答案

∑r=2n1r(r−1)=1−1n\displaystyle\sum_{r=2}^{n} \frac{1}{r(r-1)} = 1 - \frac{1}{n} (Shown)

3(b)

1. Separate the sum into two individual summations and substitute the result from part (a).

∑r=2n(12r+1r(r−1))=∑r=2n12r+∑r=2n1r(r−1)\sum_{r=2}^{n} \left( \frac{1}{2^r} + \frac{1}{r(r-1)} \right) = \sum_{r=2}^{n} \frac{1}{2^r} + \sum_{r=2}^{n} \frac{1}{r(r-1)}
Using the result from part (a):
∑r=2n1r(r−1)=1−1n\sum_{r=2}^{n} \frac{1}{r(r-1)} = 1 - \frac{1}{n}

2. Evaluate the first summation as a geometric series.

The series ∑r=2n12r\sum_{r=2}^{n} \frac{1}{2^r} is a finite geometric progression.

  • First term, a=122=14a = \frac{1}{2^2} = \frac{1}{4}
  • Common ratio, r=12r = \frac{1}{2}
  • Number of terms, N=n−2+1=n−1N = n - 2 + 1 = n - 1

Applying the sum formula SN=a(1−rN)1−rS_N = \frac{a(1 - r^N)}{1 - r}:
∑r=2n12r=14(1−(12)n−1)1−12\sum_{r=2}^{n} \frac{1}{2^r} = \frac{\frac{1}{4} \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)}{1 - \frac{1}{2}}
=14(1−(12)n−1)12= \frac{\frac{1}{4} \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)}{\frac{1}{2}}
=12(1−(12)n−1)= \frac{1}{2} \left( 1 - \left(\frac{1}{2}\right)^{n-1} \right)
=12−12(12)n−1=12−(12)n= \frac{1}{2} - \frac{1}{2} \left(\frac{1}{2}\right)^{n-1} = \frac{1}{2} - \left(\frac{1}{2}\right)^n

3. Combine the evaluated parts and simplify.

∑r=2n(12r+1r(r−1))=(12−(12)n)+(1−1n)\sum_{r=2}^{n} \left( \frac{1}{2^r} + \frac{1}{r(r-1)} \right) = \left( \frac{1}{2} - \left(\frac{1}{2}\right)^n \right) + \left( 1 - \frac{1}{n} \right)
=32−(12)n−1n= \frac{3}{2} - \left(\frac{1}{2}\right)^n - \frac{1}{n}

答案

\frac{3}{2} - \left(\frac{1}{2}\right)^n - \frac{1}{n}

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