特殊算式求正平方根

n是自然數,因為

1+$(\dfrac{1}{n})^2+(\dfrac{1}{(n+1)})^2$=1+$\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}$=

1+$\dfrac{(n+1)^2+n^2}{n^2(n+1)^2}$=1+$\dfrac{2n(n+1)+1}{n^2(n+1)^2}$=1+$\dfrac{2n(n+1)}{n^2(n+1)^2}$+$\dfrac{1}{n^2(n+1)^2}$=1+$\dfrac{2}{n(n+1)}$+$(\dfrac{1}{n(n+1)})^2$=

(1+$\dfrac{1}{n(n+1)}$)2

所以$\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}}$  =1+$\dfrac{1}{n(n+1)}$

例如

$\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{(100)^2}}$  =1+$\dfrac{1}{9900}$ =$\dfrac{9901}{9900}$

$\sqrt{1+\dfrac{1}{100^2}+\dfrac{1}{(101)^2}}$  =1+$\dfrac{1}{10100}$ =$\dfrac{10101}{10100}$

 

 

n= ,試求$\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}}$  =?

    

答案
 


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