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Question: Evaluate $\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt...

Evaluate ∑n=19910+n∑n=19910−n\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}.

Answer

2+1\sqrt{2}+1

Explanation

Solution

Let S1=∑n=19910+nS_1 = \sum_{n=1}^{99} \sqrt{10 + \sqrt{n}} and S2=∑n=19910−nS_2 = \sum_{n=1}^{99} \sqrt{10 - \sqrt{n}}. We use the property ∑k=1Nf(k)=∑k=1Nf(N+1−k)\sum_{k=1}^{N} f(k) = \sum_{k=1}^{N} f(N+1-k). So S1=∑n=19910+100−nS_1 = \sum_{n=1}^{99} \sqrt{10 + \sqrt{100-n}}. S2=∑n=19910−100−nS_2 = \sum_{n=1}^{99} \sqrt{10 - \sqrt{100-n}}.

Now we consider S1+S2S_1+S_2. S1+S2=∑n=199(10+100−n+10−100−n)S_1+S_2 = \sum_{n=1}^{99} (\sqrt{10 + \sqrt{100-n}} + \sqrt{10 - \sqrt{100-n}}). Let A=10A = 10, B=100−nB = \sqrt{100-n}. The term is A+B+A−B\sqrt{A+B} + \sqrt{A-B}. We know (A+B+A−B)2=(A+B)+(A−B)+2(A+B)(A−B)=2A+2A2−B2(\sqrt{A+B} + \sqrt{A-B})^2 = (A+B) + (A-B) + 2\sqrt{(A+B)(A-B)} = 2A + 2\sqrt{A^2-B^2}. So A+B+A−B=2A+2A2−B2\sqrt{A+B} + \sqrt{A-B} = \sqrt{2A + 2\sqrt{A^2-B^2}}. Substitute A=10A=10 and B=100−nB=\sqrt{100-n}. A2=100A^2 = 100. B2=100−nB^2 = 100-n. A2−B2=100−(100−n)=nA^2-B^2 = 100 - (100-n) = n. So the term is 2(10)+2n=20+2n=2(10+n)\sqrt{2(10) + 2\sqrt{n}} = \sqrt{20 + 2\sqrt{n}} = \sqrt{2(10 + \sqrt{n})}. This is equal to 210+n\sqrt{2} \sqrt{10 + \sqrt{n}}.

So S1+S2=∑n=199210+n=2∑n=19910+n=2S1S_1+S_2 = \sum_{n=1}^{99} \sqrt{2} \sqrt{10 + \sqrt{n}} = \sqrt{2} \sum_{n=1}^{99} \sqrt{10 + \sqrt{n}} = \sqrt{2} S_1. S1+S2=2S1S_1+S_2 = \sqrt{2}S_1. S2=(2−1)S1S_2 = (\sqrt{2}-1)S_1. Therefore, S1S2=12−1=2+1(2−1)(2+1)=2+12−1=2+1\frac{S_1}{S_2} = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1.

Thus, \frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}} = \sqrt{2}+1.