Đáp án: $I=\dfrac{11}{20}$
Giải thích các bước giải:
Ta có:
$I=(1-\dfrac14)(1-\dfrac19)(1-\dfrac1{16})....(1-\dfrac1{100})$
$\to I=(1-\dfrac1{2^2})(1-\dfrac1{3^2})(1-\dfrac1{4^2})....(1-\dfrac1{10^2})$
$\to I=\dfrac{2^2-1}{2^2}\cdot \dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}....\dfrac{10^2-1}{10^2}$
$\to I=\dfrac{(2-1)(2+1)}{2^2}\cdot \dfrac{(3-1)(3+1)}{3^2}\cdot\dfrac{(4-1)(4+1)}{4^2}....\dfrac{(10-1)(10+1)}{10^2}$
$\to I=\dfrac{1\cdot 3}{2^2}\cdot \dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot 5}{4^2}....\dfrac{9\cdot11}{10^2}$
$\to I=\dfrac{1}{2}\cdot \dfrac{11}{10}$
$\to I=\dfrac{11}{20}$
