Question

[Sand Cube]

Tính P = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\sqrt{1+\frac{1}{4^2}+\frac{1}{5^2}}+...+\sqrt{1+\frac{1}{2016^2}+\frac{1}{2017^2}}\)

Answer 1

[shitbo]

\(Tongquat:\)

\(\sqrt{1+\frac{1}{n}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n}+\frac{2}{n}-\frac{2}{n+1}-\frac{2}{n\left(n+1\right)}+\frac{1}{\left(n+1\right)^2}}\)

\(=\sqrt{\left(1+\frac{1}{n}\right)^2-2\left(1+\frac{1}{n}\right)\frac{1}{n+1}+\frac{1}{n+1}}=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}\)

\(=|1+\frac{1}{n}-\frac{1}{n+1}|=1+\frac{1}{n}-\frac{1}{n+1}\)

Thay vào ta có:

\(P=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+.........-\frac{1}{2017}\)

\(P=2015+\frac{1}{2}-\frac{1}{2017}=2015+\frac{2015}{4034}\)