\sqrt(1+(1)/(1^(2))+(1)/(2^(2)))+\sqrt(1+(1)/(2^(2))+(1)/(3^(2)))+...+\sqrt(1+(1)/(99^(2))+(1)/(100^(2)))

Đáp án:

$\dfrac{9999}{100}$. 

Giải thích các bước giải:

$\sqrt{1+\dfrac1{1^2}+\dfrac1{2^2}}+\sqrt{1+\dfrac1{2^2}+\dfrac1{3^2}}\ +\,.\!.\!.+\ \sqrt{1+\dfrac1{99^2}+\dfrac1{100^2}}$

$=\sqrt{\dfrac{1^2.2^2+1^2+2^2}{1^2.2^2}}+\sqrt{\dfrac{2^2.3^2+2^2+3^2}{2^2.3^2}}\ +\,.\!.\!.+\ \sqrt{\dfrac{99^2.100^2+99^2+100^2}{99^2.100^2}}$

$=\sqrt{\dfrac9{1^2.2^2}}+\sqrt{\dfrac{49}{2^2.3^2}}\ +\,.\!.\!.+\ \sqrt{\dfrac{99^2.100^2+99^2+1+100^2-1}{99^2.100^2}}$

$=\sqrt{\dfrac{3^2}{1^2.2^2}}+\sqrt{\dfrac{7^2}{2^2.3^2}}\ +\,.\!.\!.+\ \sqrt{\dfrac{99^2.100^2+99^2+1+(100-1)(100+1)}{99^2.100^2}}$

$=\dfrac3{1.2}+\dfrac7{2.3}\ +\,.\!.\!.+\ \dfrac{\sqrt{99^2.100^2+99^2+1+99.101}}{99.100}$

$=\dfrac3{1.2}+\dfrac7{2.3}\ +\,.\!.\!.+\ \dfrac{\sqrt{99^2.100^2+1+99.(99+101)}}{99.100}$

$=\dfrac3{1.2}+\dfrac7{2.3}\ +\,.\!.\!.+\ \dfrac{\sqrt{99^2.100^2+1+99.200}}{99.100}$

$=\dfrac3{1.2}+\dfrac7{2.3}\ +\,.\!.\!.+\ \dfrac{\sqrt{99^2.100^2+2.99.100+1}}{99.100}$

$=\dfrac3{1.2}+\dfrac7{2.3}\ +\,.\!.\!.+\ \dfrac{\sqrt{\left(99.100+1\right)^2}}{99.100}$

$=\dfrac{1.2+1}{1.2}+\dfrac{2.3+1}{2.3}\ +\,.\!.\!.+\ \dfrac{99.100+1}{99.100}$

$=1+\dfrac1{1.2}+1+\dfrac1{2.3}\ +\,.\!.\!.+\ 1+\dfrac1{99.100}$

$=99+\dfrac1{1.2}+\dfrac1{2.3}\ +\,.\!.\!.+\ \dfrac1{99.100}$

$=99+1-\dfrac12+\dfrac12-\dfrac13\ +\,.\!.\!.+\ \dfrac1{99}-\dfrac1{100}$

$=99+1-\dfrac1{100}=100-\dfrac1{100}=\dfrac{9999}{100}$