ta xét $ k\in N ,k\geq2$có : $(1+\frac{1}{k - 1} - \frac{1}{k})^{2} = 1+ \frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+\frac{2}{k -1}- \frac{2}{k(k-1)}- \frac{2}{k}$ $= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+ \frac{2}{k- 1} -\frac{2}{k -1}+ \frac{2}{k} -\frac{2}{k}= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}$ $\Rightarrow \sqrt{\frac{1}{1^{2}} +\frac{1}{(k-1)^{2}} +\frac{1}{k^{2}}} = 1+ \frac{1}{k -1}- \frac{1}{k}$ khi đó S= $ 1 +\frac{1}{1}- \frac{1}{2}+1+ \frac{1}{2} -\frac{1}{3}+...+1+ \frac{1}{99} -\frac{1}{100}$ =100- $\frac{1}{100}$
ta xét $ k\in N ,k\geq2$có : $(1+\frac{1}{k - 1} - \frac{1}{k})^{2} = 1+ \frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+\frac{2}{k -1}- \frac{2}{k(k-1)}- \frac{2}{k}$ $= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+ \frac{2}{k- 1} -\frac{2}{k -1}+ \frac{2}{k} -\frac{2}{k}= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}$ $\Rightarrow \sqrt{\frac{1}{1^{2}} +\frac{1}{(k-1)^{2}} +\frac{1}{k^{2}}} = 1+ \frac{1}{k -1}- \frac{1}{k}$ khi đó S= $ 1 +\frac{1}{1}- \frac{1}{2}+1+ \frac{1}{2} -\frac{1}{3}+...+1+ \frac{1}{99} -\frac{1}{100}$ =99- $\frac{1}{100}$
ta xét $ k\in N ,k\geq2$có : $(1+\frac{1}{k - 1} - \frac{1}{k})^{2} = 1+ \frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+\frac{2}{k -1}- \frac{2}{k(k-1)}- \frac{2}{k}$ $= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+ \frac{2}{k- 1} -\frac{2}{k -1}+ \frac{2}{k} -\frac{2}{k}= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}$ $\Rightarrow \sqrt{\frac{1}{1^{2}} +\frac{1}{(k-1)^{2}} +\frac{1}{k^{2}}} = 1+ \frac{1}{k -1}- \frac{1}{k}$ khi đó S= $ 1 +\frac{1}{1}- \frac{1}{2}+1+ \frac{1}{2} -\frac{1}{3}+...+1+ \frac{1}{99} -\frac{1}{100}$ =
100- $\frac{1}{100}$