Đáp án: $B=\dfrac{99}{50}$
Giải thích các bước giải:
Ta có :
$B=1\dfrac13.1\dfrac18.1\dfrac1{15}...$
$\to B=\left(1+\dfrac13\right).\left(1+\dfrac18\right).\left(1+\dfrac1{15}\right)...$
$\to B=\left(1+\dfrac1{2^2-1}\right).\left(1+\dfrac1{3^2-1}\right).\left(1+\dfrac1{4^2-1}\right)...$
$\to B=\left(1+\dfrac1{2^2-1}\right).\left(1+\dfrac1{3^2-1}\right).\left(1+\dfrac1{4^2-1}\right)...\left(1+\dfrac1{99^2-1}\right)$
$\to B=\left(\dfrac{2^2-1+1}{2^2-1}\right).\left(\dfrac{3^2-1+1}{3^2-1}\right).\left(\dfrac{4^2-1+1}{4^2-1}\right)...\left(\dfrac{99^2-1+1}{99^2-1}\right)$
$\to B=\left(\dfrac{2^2}{2^2-1}\right).\left(\dfrac{3^2}{3^2-1}\right).\left(\dfrac{4^2}{4^2-1}\right)...\left(\dfrac{99^2}{99^2-1}\right)$
$\to B=\dfrac{2^2}{2^2-1}.\dfrac{3^2}{3^2-1}.\dfrac{4^2}{4^2-1}...\dfrac{99^2}{99^2-1}$
$\to B=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}.\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}.\dfrac{4^2}{\left(4-1\right)\left(4+1\right)}...\dfrac{99^2}{\left(99-1\right)\left(99+1\right)}$
$\to B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{99^2}{98.100}$
$\to B=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}...\dfrac{99.99}{98.100}$
$\to B=\dfrac{2.99}{1.100}$
$\to B=\dfrac{99}{50}$
