\(B=\left(\dfrac{1}{3}-1\right)+\left(\dfrac{1}{6}-1\right)+\left(\dfrac{1}{10}-1\right)+...+\left(\dfrac{1}{45}-1\right)\)
Quy luật dãy số \(B=\left(\dfrac{1}{1+2}-1\right)+\left(\dfrac{1}{1+2+3}-1\right)+\left(\dfrac{1}{1+2+3+4}-1\right)+...+\left(\dfrac{1}{1+2+..+9}-1\right)\)\(B=\left(\dfrac{1}{1+2}\right)+\left(\dfrac{1}{1+2+3}\right)+\left(\dfrac{1}{1+2+3+4}\right)+...+\left(\dfrac{1}{1+2+..+9}\right)-8\)\(B=B_1-8\)
\(\dfrac{B_1}{2}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}\)
\(B_1=1-\dfrac{1}{5}\)
\(B=1-\dfrac{1}{5}-8=-8-\dfrac{1}{5}=-\dfrac{41}{5}\)
