The discussion focuses on calculating the value of A, defined as the sum of a series of square root expressions involving fractions of integers. The corrected formula for A is given as $A=\sqrt{1^2+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1^2+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1^2+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\ldots+\sqrt{1^2+\dfrac{1}{2011^2}+\dfrac{1}{2012^2}}$. The goal is to find an integer S that is closest to A but less than A. The final answer provided in the discussion confirms the correctness of the calculations performed.
PREREQUISITESMathematicians, students studying calculus or numerical methods, and anyone interested in series summation and integer approximation techniques.
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:$$A=\sum_{n=1}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sum_{n=1}^{2011}\sqrt{\frac{[n(n+1)]^2+(n+1)^2+n^2}{[n(n+1)]^2}}$$
$$=\sum_{n=1}^{2011}\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\sqrt{\frac{(n^2+n+1)^2}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\frac{n^2+n+1}{n(n+1)}$$
$$=\sum_{n=1}^{2011}\left(1+\frac{1}{n(n+1)}\right)=2011+\sum_{n=1}^{2011}\left(\frac1n-\frac{1}{n+1}\right)\Rightarrow S=2011$$