เเต่ละพจน์อยู่ในรูป
$\frac{6(2n+1)}{n(n+1)(2n+1)}$
$=6(\frac{1}{n}-\frac{1}{n+1})$
จะได้ $a_{100}=6(\frac{1}{100}-\frac{1}{101})=\frac{6}{101} $
$\sum_{n = 1}^{99}a_n=6(\frac{1}{1} -\frac{1}{2} +\frac{1}{2} -\frac{1}{3} +...+\frac{1}{99} -\frac{1}{100} ) $
$=6(\frac{99}{100} )$
ดังนั้น $a_{100}-\sum_{n = 1}^{99}a_n=\frac{6}{101}-6(\frac{99}{100} )=-5.88$