$\because 1\Delta 2 = 1+2-\frac{1}{1\times 2} = 3 - (\frac{1}{1} - \frac{1}{2})$
$\because 2\Delta 3 = 2+3-\frac{1}{2\times 3} = 5 - (\frac{1}{2} - \frac{1}{3})$
$\because 3\Delta 4 = 3+4-\frac{1}{3\times 4} = 7 - (\frac{1}{3} - \frac{1}{4})$
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$\because 98\Delta 99 = 98+99-\frac{1}{98\times 99} = 197 - (\frac{1}{98} - \frac{1}{99})$
$\because 99\Delta 100 = 99+100-\frac{1}{99\times 100} = 199 - (\frac{1}{99} - \frac{1}{100})$
ดังนั้น $(1\Delta 2) +(2\Delta 3) +(3\Delta 4) + (4\Delta 5)+...+(98\Delta 99)+(99\Delta 100)$
$ = (3+5+7+....+199) - (\frac{1}{1} - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) - (\frac{1}{3} - \frac{1}{4}) - ....- (\frac{1}{98} - \frac{1}{99}) - (\frac{1}{99} - \frac{1}{100}) $
$ = [(\frac{(1+\color{red}{1}99)^2}{4}) - 1] - [(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ....+ (\frac{1}{98} - \frac{1}{99}) + (\frac{1}{99} - \frac{1}{100}) ]$
<--- สะเพร่าจนได้ ตกเลข 1 
$[10000-1]-[\frac{1}{1} - \frac{1}{100}]$
$9999 - \frac{99}{100}$