数列 u0,u1,u2,…u_0, u_1, u_2, \dots 满足 u0=1u_0 = 1 且 un=un−1+2n+1n(n+1),for all n∈N+.u_n = u_{n-1} + 2n + \frac{1}{n(n+1)}, \quad\text{for all } n \in \mathbb{N}^+.

  1. (i)

    将 1r(r+1)\dfrac{1}{r(r+1)} 表示为部分分式。由此用 nn 表示 ∑r=1n(2r+1r(r+1))\displaystyle \sum_{r=1}^n\left(2r + \dfrac{1}{r(r+1)}\right)。

  2. (ii)

    通过考虑 ur−ur−1u_r - u_{r-1} 并利用第 (i) 部分的结果，证明

    un=n(n+1)+2−1n+1,where n∈N+.u_n = n(n+1) + 2 - \dfrac{1}{n+1}, \quad\text{where } n \in \mathbb{N}^+.

  3. (iii)

    利用第 (i) 部分的结果，用 nn 表示 ∑r=1n−1(2(r+1)+1(r+1)(r+2))\displaystyle \sum_{r=1}^{n-1}\left(2(r+1) + \dfrac{1}{(r+1)(r+2)}\right)。

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完整解法

2(i)

1. Express the rational function in partial fractions.

Let 1r(r+1)=Ar+Br+1\dfrac{1}{r(r+1)} = \dfrac{A}{r} + \dfrac{B}{r+1}.

1=A(r+1)+Br1 = A(r+1) + Br

Substituting r=0r = 0, we get A=1A = 1.
Substituting r=−1r = -1, we get B=−1B = -1.

Therefore, 1r(r+1)=1r−1r+1\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}.

2. Split the summation and evaluate the arithmetic sequence portion.

∑r=1n(2r+1r(r+1))=∑r=1n2r+∑r=1n(1r−1r+1)\displaystyle \sum_{r=1}^n\left(2r + \dfrac{1}{r(r+1)}\right) = \sum_{r=1}^n 2r + \sum_{r=1}^n \left(\dfrac{1}{r} - \dfrac{1}{r+1}\right)

For the first sum, it is an arithmetic progression. Using the sum formula for an AP or standard sum formula:
∑r=1n2r=2(n(n+1)2)=n(n+1)\displaystyle \sum_{r=1}^n 2r = 2 \left( \dfrac{n(n+1)}{2} \right) = n(n+1)

3. Evaluate the second summation using the method of differences.

∑r=1n(1r−1r+1)=(1−12)+(12−13)+⋯+(1n−1n+1)\displaystyle \sum_{r=1}^n \left(\dfrac{1}{r} - \dfrac{1}{r+1}\right) = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \dots + \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right)

Since intermediate terms cancel out in pairs, we are left with the first and last terms:
=1−1n+1= 1 - \dfrac{1}{n+1}

4. Combine the evaluated parts to reach the final expression.

Total sum =n(n+1)+1−1n+1= n(n+1) + 1 - \dfrac{1}{n+1}

答案

1r−1r+1\dfrac{1}{r} - \dfrac{1}{r+1}; n(n+1)+1−1n+1\quad n(n+1) + 1 - \dfrac{1}{n+1}

2(ii)

1. Concisely obtain the sum evaluated in part (i), as it is required for the target.

From part (i), we use the partial fractions 1r(r+1)=1r−1r+1\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1} to evaluate the sum:
∑r=1n(2r+1r(r+1))=∑r=1n2r+∑r=1n(1r−1r+1)\sum_{r=1}^n\left(2r + \dfrac{1}{r(r+1)}\right) = \sum_{r=1}^n 2r + \sum_{r=1}^n \left(\dfrac{1}{r} - \dfrac{1}{r+1}\right)
=2(n(n+1)2)+((1−12)+(12−13)+⋯+(1n−1n+1))= 2\left(\dfrac{n(n+1)}{2}\right) + \left( (1 - \dfrac{1}{2}) + (\dfrac{1}{2} - \dfrac{1}{3}) + \dots + (\dfrac{1}{n} - \dfrac{1}{n+1}) \right)
=n(n+1)+1−1n+1= n(n+1) + 1 - \dfrac{1}{n+1}

2. Rearrange the given recurrence relation to express $u_r - u_{r-1}$ and sum both sides from $r=1$ to $n$.

From the given recurrence relation, replace nn with rr and rearrange:
ur=ur−1+2r+1r(r+1)u_r = u_{r-1} + 2r + \dfrac{1}{r(r+1)}
ur−ur−1=2r+1r(r+1)u_r - u_{r-1} = 2r + \dfrac{1}{r(r+1)}
Taking the sum from r=1r=1 to nn on both sides:
∑r=1n(ur−ur−1)=∑r=1n(2r+1r(r+1))\sum_{r=1}^n (u_r - u_{r-1}) = \sum_{r=1}^n \left(2r + \dfrac{1}{r(r+1)}\right)

3. Evaluate the left-hand side using the Method of Differences.

∑r=1n(ur−ur−1)=(u1−u0)+(u2−u1)+⋯+(un−un−1)\sum_{r=1}^n (u_r - u_{r-1}) = (u_1 - u_0) + (u_2 - u_1) + \dots + (u_n - u_{n-1})
Notice that all intermediate terms cancel out, leaving only the first and last terms:
∑r=1n(ur−ur−1)=un−u0\sum_{r=1}^n (u_r - u_{r-1}) = u_n - u_0

4. Substitute the given value of $u_0$ and the evaluated sum from part (i) to complete the proof.

Equating the evaluated left side and right side sums:
un−u0=n(n+1)+1−1n+1u_n - u_0 = n(n+1) + 1 - \dfrac{1}{n+1}
Substitute the given initial condition u0=1u_0 = 1:
un−1=n(n+1)+1−1n+1u_n - 1 = n(n+1) + 1 - \dfrac{1}{n+1}
un=n(n+1)+2−1n+1(shown)u_n = n(n+1) + 2 - \dfrac{1}{n+1} \quad \text{(shown)}

答案

un=n(n+1)+2−1n+1(shown)u_n = n(n+1) + 2 - \dfrac{1}{n+1} \quad \text{(shown)}

2(iii)

1. Evaluate the result of the sum from part (i) as it is required for this part.

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

2. Rewrite the target sum by shifting the index to relate it to the sum in part (i).

Let k=r+1k = r + 1. When r=1r=1, k=2k=2; when r=n−1r=n-1, k=nk=n.
∑r=1n−1(2(r+1)+1(r+1)(r+2))=∑k=2n(2k+1k(k+1))\sum_{r=1}^{n-1}\left(2(r+1) + \frac{1}{(r+1)(r+2)}\right) = \sum_{k=2}^n \left(2k + \frac{1}{k(k+1)}\right)
=∑k=1n(2k+1k(k+1))−(2(1)+11(2))= \sum_{k=1}^n \left(2k + \frac{1}{k(k+1)}\right) - \left( 2(1) + \frac{1}{1(2)} \right)

3. Substitute the evaluated sum from part (i) and simplify.

=(n(n+1)+1−1n+1)−(2+12)= \left( n(n+1) + 1 - \frac{1}{n+1} \right) - \left( 2 + \frac{1}{2} \right)
=n2+n+1−1n+1−52= n^2 + n + 1 - \frac{1}{n+1} - \frac{5}{2}
=n2+n−32−1n+1= n^2 + n - \frac{3}{2} - \frac{1}{n+1}

答案

n2+n−32−1n+1n^2 + n - \frac{3}{2} - \frac{1}{n+1}

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