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4

Part of 2007 Greece JBMO TST

Problems(1)

sum with last term \sqrt{1+(8\cdot 1003^2-1)/(2005^2 2007^2)}

Source: Greece JBMO TST 2007 p4

6/17/2019
Calculate the sum S=1+8⋅12−112⋅32+1+8⋅22−132⋅52+...+1+8⋅10032−120052⋅20072S=\sqrt{1+\frac{8\cdot 1^2-1}{1^2\cdot 3^2}}+\sqrt{1+\frac{8\cdot 2^2-1}{3^2\cdot 5^2}}+...+ \sqrt{1+\frac{8\cdot 1003^2-1}{2005^2\cdot 2007^2}}
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