Division

Algebra Level 5

( ∑ n = 1 99 10 + n ) ÷ ( ∑ n = 1 99 10 − n ) {\displaystyle \left(\sum _{n=1}^{99}\sqrt{10+\sqrt{n}} \right)} \div { \displaystyle \left(\sum _{n=1}^{99}\sqrt{10-\sqrt{n}}\right) }

If the expression above is equal to a + b c a+b\sqrt{c} for integers a , b , c a,b,c with c c square-free, find a + b + c a+b+c .


The answer is 4.

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1 solution

Ravi Dwivedi
Jan 10, 2016

Let A = ( ∑ n = 1 99 10 + n ) A= {\displaystyle \left(\sum _{n=1}^{99}\sqrt{10+\sqrt{n}} \right)}

B = ( ∑ n = 1 99 10 − n ) B= { \displaystyle \left(\sum _{n=1}^{99}\sqrt{10-\sqrt{n}}\right) }

We have:

A + B = ∑ i = 1 99 ( 10 + 100 − i + 10 − 100 − i ) = ∑ i = 1 99 ( 10 + 100 − i + 10 − 100 − i ) 2 = ∑ i = 1 99 20 + 2 i = 2 A \displaystyle A+B=\sum_{i=1}^{99}\left(\sqrt{10+\sqrt{100-i}}+\sqrt{10-\sqrt{100-i}}\right)\\ \qquad\;\;\displaystyle=\sum_{i=1}^{99}\sqrt{\left(\sqrt{10+\sqrt{100-i}}+\sqrt{10-\sqrt{100-i}}\right)^2}\\ \qquad\;\;\displaystyle =\sum_{i=1}^{99}\sqrt{20+2\sqrt{i}}=\sqrt{2}A

Thus, B = ( 2 − 1 ) A B=(\sqrt{2}-1)A or A B = 1 2 − 1 = 2 + 1 \dfrac{A}{B}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1

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What makes you think of showing A + B = 2 A A + B = \sqrt{2} A ?

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