证明 1a+b=a−ba−b\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{a-b}。 精确求出 ∑r=1250014r−3+4r+1\displaystyle\sum_{r=1}^{2500}\dfrac{1}{\sqrt{4r-3}+\sqrt{4r+1}} 的值。 由此证明，在不使用计算器的情况下，∑r=0250014r+1+4r+5>24\displaystyle\sum_{r=0}^{2500}\dfrac{1}{\sqrt{4r+1}+\sqrt{4r+5}}>24。

裂项相消求和与差分通项公式English
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完整解法

10((i))

1. Multiply the numerator and the denominator by the conjugate $\sqrt{a}-\sqrt{b}$.

1a+b=1a+b×a−ba−b\begin{align*}\frac{1}{\sqrt{a}+\sqrt{b}} &= \frac{1}{\sqrt{a}+\sqrt{b}} \times \frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}\end{align*}

2. Simplify the denominator using the difference of squares identity.

=a−b(a)2−(b)2=a−ba−b\begin{align*} &= \frac{\sqrt{a}-\sqrt{b}}{(\sqrt{a})^2-(\sqrt{b})^2} \\ &= \frac{\sqrt{a}-\sqrt{b}}{a-b}\end{align*}
答案

a−ba−b\dfrac{\sqrt{a}-\sqrt{b}}{a-b} (Shown)

10((ii))

1. Rationalise the denominator of the general term using the identity proven in part (i).

Let a=4r−3a = 4r-3 and b=4r+1b = 4r+1. Using the result from part (i):
14r−3+4r+1=4r−3−4r+1(4r−3)−(4r+1)\frac{1}{\sqrt{4r-3}+\sqrt{4r+1}} = \frac{\sqrt{4r-3}-\sqrt{4r+1}}{(4r-3)-(4r+1)}
=4r−3−4r+1−4= \frac{\sqrt{4r-3}-\sqrt{4r+1}}{-4}
=14(4r+1−4r−3)= \frac{1}{4}\left(\sqrt{4r+1}-\sqrt{4r-3}\right)

2. Write out the first few and last terms of the series to reveal the telescoping cancellation.

∑r=1250014r−3+4r+1=14∑r=12500(4r+1−4r−3)\sum_{r=1}^{2500}\frac{1}{\sqrt{4r-3}+\sqrt{4r+1}} = \frac{1}{4} \sum_{r=1}^{2500} \left( \sqrt{4r+1} - \sqrt{4r-3} \right)
=14[(5−1)+(9−5)+(13−9)+⋯+(10001−9997)]= \frac{1}{4} \left[ (\sqrt{5} - \sqrt{1}) + (\sqrt{9} - \sqrt{5}) + (\sqrt{13} - \sqrt{9}) + \dots + (\sqrt{10001} - \sqrt{9997}) \right]

3. Cancel the intermediate terms to find the exact sum.

All intermediate terms cancel out (method of differences), leaving only the negative of the first term and the positive of the last term:
=14(−1+10001)= \frac{1}{4} (-\sqrt{1} + \sqrt{10001})
=10001−14= \frac{\sqrt{10001} - 1}{4}

答案

10001−14\dfrac{\sqrt{10001}-1}{4}

10((iii))

1. Apply the identity from part (i) to express the given sum as a series of differences.

Let S=∑r=0250014r+1+4r+5\text{Let } S = \sum_{r=0}^{2500}\dfrac{1}{\sqrt{4r+1}+\sqrt{4r+5}}\nFrom part (i), replacing aa with 4r+14r+1 and bb with 4r+54r+5:\n14r+1+4r+5=4r+1−4r+5(4r+1)−(4r+5)\dfrac{1}{\sqrt{4r+1}+\sqrt{4r+5}} = \dfrac{\sqrt{4r+1}-\sqrt{4r+5}}{(4r+1)-(4r+5)}\n=4r+1−4r+5−4=4r+5−4r+14= \dfrac{\sqrt{4r+1}-\sqrt{4r+5}}{-4} = \dfrac{\sqrt{4r+5}-\sqrt{4r+1}}{4}

2. Expand the sum to observe cancellation of intermediate terms and find its exact value.

S=14∑r=02500(4r+5−4r+1)S = \dfrac{1}{4} \sum_{r=0}^{2500} (\sqrt{4r+5} - \sqrt{4r+1})\nS=14[(5−1)+(9−5)+⋯+(10001−9997)+(10005−10001)]S = \dfrac{1}{4} \left[ (\sqrt{5} - \sqrt{1}) + (\sqrt{9} - \sqrt{5}) + \dots + (\sqrt{10001} - \sqrt{9997}) + (\sqrt{10005} - \sqrt{10001}) \right]\nBy cancelling intermediate terms, we get:\nS=14(10005−1)S = \dfrac{1}{4} (\sqrt{10005} - 1)

3. Establish the inequality by bounding the exact sum from below using a known perfect square.

Since 10005>10000, it follows that 10005>10000=100.\text{Since } 10005 > 10000, \text{ it follows that } \sqrt{10005} > \sqrt{10000} = 100.\n∴S>14(100−1)=994=24.75\therefore S > \dfrac{1}{4} (100 - 1) = \dfrac{99}{4} = 24.75\nSince 24.75>24, we have shown that ∑r=0250014r+1+4r+5>24.\text{Since } 24.75 > 24, \text{ we have shown that } \sum_{r=0}^{2500}\dfrac{1}{\sqrt{4r+1}+\sqrt{4r+5}} > 24.

答案

Shown\text{Shown}

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