The discussion revolves around finding an integer S that is closest to a defined value A, while also being less than A. The value A is expressed as a summation involving square roots and fractions, and the problem is presented in a mathematical context.
The discussion does not present any explicit disagreements, but it does not clarify whether there is consensus on the value of S or the correctness of the calculations leading to it.
The discussion involves a mathematical expression that may have dependencies on specific interpretations of the summation and the calculations involved, which are not fully resolved in the posts.
sorry :a typoAlbert said:$A=\sqrt{1^2+\dfrac{1}{1^2+2^2}}+\sqrt{1^2+\dfrac{1}{2^2+3^2}}+\sqrt{1^2+\dfrac{1}{3^2+4^2}}+---+\sqrt{1^2+\dfrac{1}{2011^2+2012^2}}$
please find an integer S most close to A and less than A
great ! your answer is correctgreg1313 said:$$A=\sum_{n=1}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sum_{n=1}^{2011}\sqrt{\frac{[n(n+1)]^2+(n+1)^2+n^2}{[n(n+1)]^2}}$$
$$=\sum_{n=1}^{2011}\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\sqrt{\frac{(n^2+n+1)^2}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\frac{n^2+n+1}{n(n+1)}$$
$$=\sum_{n=1}^{2011}\left(1+\frac{1}{n(n+1)}\right)=2011+\sum_{n=1}^{2011}\left(\frac1n-\frac{1}{n+1}\right)\Rightarrow S=2011$$