Toán 9 Tính

Nguyễn Cao Trường said:

[tex]\sqrt{1+1+1/2^2}+\sqrt{1+1/2^2+1/3^2}+...+\sqrt{1+1/2018^2+1/2019^2}[/tex]

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Đưa về dạng tổng quát;
[tex]\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^2}}=\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^2}+\frac{2}{n}-\frac{2}{n+1}-\frac{2}{n(n+1)}}=\sqrt{(1+\frac{1}{n}-\frac{1}{n+1})^2}[/tex]
Vậy pt trên <=> [tex]\sqrt{(1+1-\frac{1}{2})^2}+\sqrt{(1+\frac{1}{2}-\frac{1}{3})^2}+...+\sqrt{(1+\frac{1}{2018}-\frac{1}{2019})^2}[/tex]
=[tex]1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2018}-\frac{1}{2019}=2019-\frac{1}{2019}=\frac{2019^2-1}{2019}[/tex]